How should I determine all numbers $zin mathbb C$ such that $z^3 = 4overline z$?












0














Find all $zin mathbb C$ such that $z^3 = 4overline z$



I have set $z = re^{itheta}$ and found that $z^3 = 4(x-iy)$ leads to the equation $z = 2e^{-itheta}$. I am supposed to determine all numbers $z$ with that property, and have just found that result and said that it would be all numbers with a radius of $2$ and with $z$ going clockwise as the argument increases. Is there a way for me to be more thorough in answering this or am I done?










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  • 1




    Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
    – lulu
    Dec 3 '18 at 0:31
















0














Find all $zin mathbb C$ such that $z^3 = 4overline z$



I have set $z = re^{itheta}$ and found that $z^3 = 4(x-iy)$ leads to the equation $z = 2e^{-itheta}$. I am supposed to determine all numbers $z$ with that property, and have just found that result and said that it would be all numbers with a radius of $2$ and with $z$ going clockwise as the argument increases. Is there a way for me to be more thorough in answering this or am I done?










share|cite|improve this question




















  • 1




    Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
    – lulu
    Dec 3 '18 at 0:31














0












0








0







Find all $zin mathbb C$ such that $z^3 = 4overline z$



I have set $z = re^{itheta}$ and found that $z^3 = 4(x-iy)$ leads to the equation $z = 2e^{-itheta}$. I am supposed to determine all numbers $z$ with that property, and have just found that result and said that it would be all numbers with a radius of $2$ and with $z$ going clockwise as the argument increases. Is there a way for me to be more thorough in answering this or am I done?










share|cite|improve this question















Find all $zin mathbb C$ such that $z^3 = 4overline z$



I have set $z = re^{itheta}$ and found that $z^3 = 4(x-iy)$ leads to the equation $z = 2e^{-itheta}$. I am supposed to determine all numbers $z$ with that property, and have just found that result and said that it would be all numbers with a radius of $2$ and with $z$ going clockwise as the argument increases. Is there a way for me to be more thorough in answering this or am I done?







complex-numbers






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edited Dec 3 '18 at 0:33









lulu

39.2k24677




39.2k24677










asked Dec 3 '18 at 0:23









Rey

12




12








  • 1




    Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
    – lulu
    Dec 3 '18 at 0:31














  • 1




    Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
    – lulu
    Dec 3 '18 at 0:31








1




1




Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
– lulu
Dec 3 '18 at 0:31




Well, $z=0$ is an obvious solution, not of the form you wrote. Also, not every $z$ of the form you wrote is a solution. You need to say which $theta $ will work.
– lulu
Dec 3 '18 at 0:31










2 Answers
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4














There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.



Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.






share|cite|improve this answer





















  • "There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
    – fleablood
    Dec 3 '18 at 1:08



















1














choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.



You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$



or



$r^3 e^{i3theta} = 4re^{-itheta}$



So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$



$r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.



If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$



And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.



And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.



And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.



===



For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so



$x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$



$x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$



$x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$



Case 1:$x = 0$ then $-y^3 = y$



Case 1a: $x = 0$ and $y=0$ and $z = 0$.



Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.



So $z = 2i$ or $z = -2i$



Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$



Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.



So $z = 2$ or $z = -2$.



Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.



So solutions are ${0,2,2i,-2,-2i}$



that actually wasn't so bad. But usually it'd be a lot harder.



=====



I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.



And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.






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    2 Answers
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    2 Answers
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    4














    There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.



    Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.






    share|cite|improve this answer





















    • "There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
      – fleablood
      Dec 3 '18 at 1:08
















    4














    There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.



    Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.






    share|cite|improve this answer





















    • "There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
      – fleablood
      Dec 3 '18 at 1:08














    4












    4








    4






    There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.



    Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.






    share|cite|improve this answer












    There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.



    Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 0:49









    Barry Cipra

    59.1k653124




    59.1k653124












    • "There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
      – fleablood
      Dec 3 '18 at 1:08


















    • "There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
      – fleablood
      Dec 3 '18 at 1:08
















    "There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
    – fleablood
    Dec 3 '18 at 1:08




    "There is no real need to invoke polar coordinates." No reason not to either. Polar coordinates making exponents linear equations that are trivial to solve. $z^3 = 4overline z$ implies $z^4 = 4|z|^2$ implying pur real or imaginary surely isn't intuitively obvious. (But it sure is clever!) [and either way $r = |z|$ being so that $|z|^3 = 4|z|$ is equally clear either way we do it.
    – fleablood
    Dec 3 '18 at 1:08











    1














    choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.



    You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$



    or



    $r^3 e^{i3theta} = 4re^{-itheta}$



    So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$



    $r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.



    If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$



    And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.



    And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.



    And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.



    ===



    For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so



    $x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$



    $x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$



    $x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$



    Case 1:$x = 0$ then $-y^3 = y$



    Case 1a: $x = 0$ and $y=0$ and $z = 0$.



    Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.



    So $z = 2i$ or $z = -2i$



    Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$



    Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.



    So $z = 2$ or $z = -2$.



    Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.



    So solutions are ${0,2,2i,-2,-2i}$



    that actually wasn't so bad. But usually it'd be a lot harder.



    =====



    I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.



    And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.






    share|cite|improve this answer




























      1














      choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.



      You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$



      or



      $r^3 e^{i3theta} = 4re^{-itheta}$



      So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$



      $r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.



      If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$



      And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.



      And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.



      And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.



      ===



      For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so



      $x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$



      $x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$



      $x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$



      Case 1:$x = 0$ then $-y^3 = y$



      Case 1a: $x = 0$ and $y=0$ and $z = 0$.



      Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.



      So $z = 2i$ or $z = -2i$



      Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$



      Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.



      So $z = 2$ or $z = -2$.



      Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.



      So solutions are ${0,2,2i,-2,-2i}$



      that actually wasn't so bad. But usually it'd be a lot harder.



      =====



      I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.



      And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.






      share|cite|improve this answer


























        1












        1








        1






        choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.



        You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$



        or



        $r^3 e^{i3theta} = 4re^{-itheta}$



        So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$



        $r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.



        If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$



        And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.



        And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.



        And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.



        ===



        For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so



        $x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$



        $x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$



        $x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$



        Case 1:$x = 0$ then $-y^3 = y$



        Case 1a: $x = 0$ and $y=0$ and $z = 0$.



        Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.



        So $z = 2i$ or $z = -2i$



        Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$



        Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.



        So $z = 2$ or $z = -2$.



        Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.



        So solutions are ${0,2,2i,-2,-2i}$



        that actually wasn't so bad. But usually it'd be a lot harder.



        =====



        I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.



        And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.






        share|cite|improve this answer














        choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.



        You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$



        or



        $r^3 e^{i3theta} = 4re^{-itheta}$



        So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$



        $r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.



        If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$



        And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.



        And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.



        And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.



        ===



        For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so



        $x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$



        $x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$



        $x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$



        Case 1:$x = 0$ then $-y^3 = y$



        Case 1a: $x = 0$ and $y=0$ and $z = 0$.



        Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.



        So $z = 2i$ or $z = -2i$



        Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$



        Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.



        So $z = 2$ or $z = -2$.



        Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.



        So solutions are ${0,2,2i,-2,-2i}$



        that actually wasn't so bad. But usually it'd be a lot harder.



        =====



        I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.



        And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.







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        edited Dec 3 '18 at 1:12

























        answered Dec 3 '18 at 0:38









        fleablood

        68.4k22685




        68.4k22685






























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