Why is $cos(i)>1$?












8












$begingroup$


I always thought that cosine only ranges from $-1$ to $1$. But, I found out that
$$cos(i)=frac12left(e+frac{1}{e}right)$$
which is certainly greater than $1$. Why is that?






trigonometry complex-numbers







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    $-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
    $endgroup$
    – saulspatz
    Jan 5 at 17:41






  • 3




    $begingroup$
    Complex numbers typically break rules that were established within the confines of the real numbers.
    $endgroup$
    – Blue
    Jan 5 at 17:41






  • 2




    $begingroup$
    Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
    $endgroup$
    – Ben W
    Jan 5 at 17:53










  • $begingroup$
    If $cos(z)$ where bounded, it would be constant.
    $endgroup$
    – NewMath
    Jan 5 at 17:55








  • 1




    $begingroup$
    Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
    $endgroup$
    – Ennar
    Jan 5 at 18:39
















8












$begingroup$


I always thought that cosine only ranges from $-1$ to $1$. But, I found out that
$$cos(i)=frac12left(e+frac{1}{e}right)$$
which is certainly greater than $1$. Why is that?






trigonometry complex-numbers







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    $-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
    $endgroup$
    – saulspatz
    Jan 5 at 17:41






  • 3




    $begingroup$
    Complex numbers typically break rules that were established within the confines of the real numbers.
    $endgroup$
    – Blue
    Jan 5 at 17:41






  • 2




    $begingroup$
    Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
    $endgroup$
    – Ben W
    Jan 5 at 17:53










  • $begingroup$
    If $cos(z)$ where bounded, it would be constant.
    $endgroup$
    – NewMath
    Jan 5 at 17:55








  • 1




    $begingroup$
    Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
    $endgroup$
    – Ennar
    Jan 5 at 18:39














8












8








8


1



$begingroup$


I always thought that cosine only ranges from $-1$ to $1$. But, I found out that
$$cos(i)=frac12left(e+frac{1}{e}right)$$
which is certainly greater than $1$. Why is that?






trigonometry complex-numbers







share|cite|improve this question











$endgroup$




I always thought that cosine only ranges from $-1$ to $1$. But, I found out that
$$cos(i)=frac12left(e+frac{1}{e}right)$$
which is certainly greater than $1$. Why is that?






trigonometry complex-numbers




trigonometry complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 17:39









Blue

49.5k870158




49.5k870158










asked Jan 5 at 17:37









Math LoverMath Lover

17410




17410








  • 4




    $begingroup$
    $-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
    $endgroup$
    – saulspatz
    Jan 5 at 17:41






  • 3




    $begingroup$
    Complex numbers typically break rules that were established within the confines of the real numbers.
    $endgroup$
    – Blue
    Jan 5 at 17:41






  • 2




    $begingroup$
    Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
    $endgroup$
    – Ben W
    Jan 5 at 17:53










  • $begingroup$
    If $cos(z)$ where bounded, it would be constant.
    $endgroup$
    – NewMath
    Jan 5 at 17:55








  • 1




    $begingroup$
    Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
    $endgroup$
    – Ennar
    Jan 5 at 18:39














  • 4




    $begingroup$
    $-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
    $endgroup$
    – saulspatz
    Jan 5 at 17:41






  • 3




    $begingroup$
    Complex numbers typically break rules that were established within the confines of the real numbers.
    $endgroup$
    – Blue
    Jan 5 at 17:41






  • 2




    $begingroup$
    Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
    $endgroup$
    – Ben W
    Jan 5 at 17:53










  • $begingroup$
    If $cos(z)$ where bounded, it would be constant.
    $endgroup$
    – NewMath
    Jan 5 at 17:55








  • 1




    $begingroup$
    Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
    $endgroup$
    – Ennar
    Jan 5 at 18:39








4




4




$begingroup$
$-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
$endgroup$
– saulspatz
Jan 5 at 17:41




$begingroup$
$-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
$endgroup$
– saulspatz
Jan 5 at 17:41




3




3




$begingroup$
Complex numbers typically break rules that were established within the confines of the real numbers.
$endgroup$
– Blue
Jan 5 at 17:41




$begingroup$
Complex numbers typically break rules that were established within the confines of the real numbers.
$endgroup$
– Blue
Jan 5 at 17:41




2




2




$begingroup$
Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
$endgroup$
– Ben W
Jan 5 at 17:53




$begingroup$
Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
$endgroup$
– Ben W
Jan 5 at 17:53












$begingroup$
If $cos(z)$ where bounded, it would be constant.
$endgroup$
– NewMath
Jan 5 at 17:55






$begingroup$
If $cos(z)$ where bounded, it would be constant.
$endgroup$
– NewMath
Jan 5 at 17:55






1




1




$begingroup$
Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
$endgroup$
– Ennar
Jan 5 at 18:39




$begingroup$
Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
$endgroup$
– Ennar
Jan 5 at 18:39










5 Answers
5






active

oldest

votes


















6












$begingroup$

The general definition of $cos(z)$ is
$$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you prove this?
    $endgroup$
    – Math Lover
    Jan 5 at 17:47










  • $begingroup$
    Sure, do you want me to add the proof as part of the answer?
    $endgroup$
    – Larry
    Jan 5 at 17:48






  • 1




    $begingroup$
    @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
    $endgroup$
    – Blue
    Jan 5 at 17:58












  • $begingroup$
    @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
    $endgroup$
    – Benjamin Thoburn
    Jan 5 at 18:20






  • 1




    $begingroup$
    @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
    $endgroup$
    – Benjamin Thoburn
    Jan 5 at 18:29



















16












$begingroup$

It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    +1 for "There's no scandal".
    $endgroup$
    – Ennar
    Jan 5 at 17:46






  • 5




    $begingroup$
    To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
    $endgroup$
    – Martin R
    Jan 5 at 17:46





















5












$begingroup$

The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



    *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      Hint



      Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$






      share|cite|improve this answer









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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        The general definition of $cos(z)$ is
        $$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
        When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Can you prove this?
          $endgroup$
          – Math Lover
          Jan 5 at 17:47










        • $begingroup$
          Sure, do you want me to add the proof as part of the answer?
          $endgroup$
          – Larry
          Jan 5 at 17:48






        • 1




          $begingroup$
          @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
          $endgroup$
          – Blue
          Jan 5 at 17:58












        • $begingroup$
          @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
          $endgroup$
          – Benjamin Thoburn
          Jan 5 at 18:20






        • 1




          $begingroup$
          @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
          $endgroup$
          – Benjamin Thoburn
          Jan 5 at 18:29
















        6












        $begingroup$

        The general definition of $cos(z)$ is
        $$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
        When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Can you prove this?
          $endgroup$
          – Math Lover
          Jan 5 at 17:47










        • $begingroup$
          Sure, do you want me to add the proof as part of the answer?
          $endgroup$
          – Larry
          Jan 5 at 17:48






        • 1




          $begingroup$
          @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
          $endgroup$
          – Blue
          Jan 5 at 17:58












        • $begingroup$
          @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
          $endgroup$
          – Benjamin Thoburn
          Jan 5 at 18:20






        • 1




          $begingroup$
          @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
          $endgroup$
          – Benjamin Thoburn
          Jan 5 at 18:29














        6












        6








        6





        $begingroup$

        The general definition of $cos(z)$ is
        $$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
        When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$






        share|cite|improve this answer









        $endgroup$



        The general definition of $cos(z)$ is
        $$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
        When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 17:44









        LarryLarry

        2,53031131




        2,53031131












        • $begingroup$
          Can you prove this?
          $endgroup$
          – Math Lover
          Jan 5 at 17:47










        • $begingroup$
          Sure, do you want me to add the proof as part of the answer?
          $endgroup$
          – Larry
          Jan 5 at 17:48






        • 1




          $begingroup$
          @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
          $endgroup$
          – Blue
          Jan 5 at 17:58












        • $begingroup$
          @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
          $endgroup$
          – Benjamin Thoburn
          Jan 5 at 18:20






        • 1




          $begingroup$
          @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
          $endgroup$
          – Benjamin Thoburn
          Jan 5 at 18:29


















        • $begingroup$
          Can you prove this?
          $endgroup$
          – Math Lover
          Jan 5 at 17:47










        • $begingroup$
          Sure, do you want me to add the proof as part of the answer?
          $endgroup$
          – Larry
          Jan 5 at 17:48






        • 1




          $begingroup$
          @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
          $endgroup$
          – Blue
          Jan 5 at 17:58












        • $begingroup$
          @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
          $endgroup$
          – Benjamin Thoburn
          Jan 5 at 18:20






        • 1




          $begingroup$
          @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
          $endgroup$
          – Benjamin Thoburn
          Jan 5 at 18:29
















        $begingroup$
        Can you prove this?
        $endgroup$
        – Math Lover
        Jan 5 at 17:47




        $begingroup$
        Can you prove this?
        $endgroup$
        – Math Lover
        Jan 5 at 17:47












        $begingroup$
        Sure, do you want me to add the proof as part of the answer?
        $endgroup$
        – Larry
        Jan 5 at 17:48




        $begingroup$
        Sure, do you want me to add the proof as part of the answer?
        $endgroup$
        – Larry
        Jan 5 at 17:48




        1




        1




        $begingroup$
        @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
        $endgroup$
        – Blue
        Jan 5 at 17:58






        $begingroup$
        @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
        $endgroup$
        – Blue
        Jan 5 at 17:58














        $begingroup$
        @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
        $endgroup$
        – Benjamin Thoburn
        Jan 5 at 18:20




        $begingroup$
        @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
        $endgroup$
        – Benjamin Thoburn
        Jan 5 at 18:20




        1




        1




        $begingroup$
        @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
        $endgroup$
        – Benjamin Thoburn
        Jan 5 at 18:29




        $begingroup$
        @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
        $endgroup$
        – Benjamin Thoburn
        Jan 5 at 18:29











        16












        $begingroup$

        It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



        Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.






        share|cite|improve this answer









        $endgroup$









        • 2




          $begingroup$
          +1 for "There's no scandal".
          $endgroup$
          – Ennar
          Jan 5 at 17:46






        • 5




          $begingroup$
          To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
          $endgroup$
          – Martin R
          Jan 5 at 17:46


















        16












        $begingroup$

        It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



        Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.






        share|cite|improve this answer









        $endgroup$









        • 2




          $begingroup$
          +1 for "There's no scandal".
          $endgroup$
          – Ennar
          Jan 5 at 17:46






        • 5




          $begingroup$
          To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
          $endgroup$
          – Martin R
          Jan 5 at 17:46
















        16












        16








        16





        $begingroup$

        It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



        Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.






        share|cite|improve this answer









        $endgroup$



        It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



        Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 17:41









        hunterhunter

        15.7k32643




        15.7k32643








        • 2




          $begingroup$
          +1 for "There's no scandal".
          $endgroup$
          – Ennar
          Jan 5 at 17:46






        • 5




          $begingroup$
          To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
          $endgroup$
          – Martin R
          Jan 5 at 17:46
















        • 2




          $begingroup$
          +1 for "There's no scandal".
          $endgroup$
          – Ennar
          Jan 5 at 17:46






        • 5




          $begingroup$
          To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
          $endgroup$
          – Martin R
          Jan 5 at 17:46










        2




        2




        $begingroup$
        +1 for "There's no scandal".
        $endgroup$
        – Ennar
        Jan 5 at 17:46




        $begingroup$
        +1 for "There's no scandal".
        $endgroup$
        – Ennar
        Jan 5 at 17:46




        5




        5




        $begingroup$
        To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
        $endgroup$
        – Martin R
        Jan 5 at 17:46






        $begingroup$
        To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
        $endgroup$
        – Martin R
        Jan 5 at 17:46













        5












        $begingroup$

        The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.






        share|cite|improve this answer











        $endgroup$


















          5












          $begingroup$

          The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.






          share|cite|improve this answer











          $endgroup$
















            5












            5








            5





            $begingroup$

            The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.






            share|cite|improve this answer











            $endgroup$



            The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 5 at 17:43









            J.G.

            33.1k23251




            33.1k23251










            answered Jan 5 at 17:42









            pendermathpendermath

            56612




            56612























                2












                $begingroup$

                Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



                *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



                  *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



                    *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).






                    share|cite|improve this answer











                    $endgroup$



                    Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



                    *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 5 at 19:49

























                    answered Jan 5 at 19:44









                    Benjamin ThoburnBenjamin Thoburn

                    363313




                    363313























                        1












                        $begingroup$

                        Hint



                        Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint



                          Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint



                            Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$






                            share|cite|improve this answer









                            $endgroup$



                            Hint



                            Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 17:49









                            Mostafa AyazMostafa Ayaz

                            18.1k31040




                            18.1k31040






























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