Some weird results in complex number computing
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The question I met is to show that if $z=cos (theta)+isin(theta)$ with $i=sqrt{-1}$,then $ Re(frac{z-1}{z+1})=0$
In the normal way, we found that: $$frac{z-1}{z+1}=frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}\
=frac{bigl(cos(theta)-1+isin(theta)bigl)bigl(cos(theta)+1-isin(theta)bigl)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{cos^2(theta)+cos(theta)-cos(theta)-1+sin^2(theta)+2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}$$
So $Re(frac{z-1}{z+1})=0$
If we do it in another way:
$$
frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}=frac{-2sin^2(frac{theta}{2})+2icos(frac{theta}{2})sin(frac{theta}{2})}{2cos^2(frac{theta}{2})+2isin(frac{theta}{2})cos(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{sin(frac{theta}{2})-icos(frac{theta}{2})}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)
$$
So the real part of it will be $-tan(frac{theta}{2})cos(-theta-frac{pi}{2})$ which is not $0$.
Which step I made mistake or they are equivalent?
complex-numbers
asked Jan 8 at 9:38
yuanming luoyuanming luo
10211
$endgroup$
|
show 1 more comment
$begingroup$
The question I met is to show that if $z=cos (theta)+isin(theta)$ with $i=sqrt{-1}$,then $ Re(frac{z-1}{z+1})=0$
In the normal way, we found that: $$frac{z-1}{z+1}=frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}\
=frac{bigl(cos(theta)-1+isin(theta)bigl)bigl(cos(theta)+1-isin(theta)bigl)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{cos^2(theta)+cos(theta)-cos(theta)-1+sin^2(theta)+2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}$$
So $Re(frac{z-1}{z+1})=0$
If we do it in another way:
$$
frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}=frac{-2sin^2(frac{theta}{2})+2icos(frac{theta}{2})sin(frac{theta}{2})}{2cos^2(frac{theta}{2})+2isin(frac{theta}{2})cos(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{sin(frac{theta}{2})-icos(frac{theta}{2})}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)
$$
So the real part of it will be $-tan(frac{theta}{2})cos(-theta-frac{pi}{2})$ which is not $0$.
Which step I made mistake or they are equivalent?
complex-numbers
asked Jan 8 at 9:38
yuanming luoyuanming luo
10211
$endgroup$
$begingroup$
There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
$endgroup$
– Yves Daoust
Jan 8 at 9:44
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Demoivre's theory, the arguments of two complex number are added.
$endgroup$
– yuanming luo
Jan 8 at 9:47
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How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
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– 5xum
Jan 8 at 9:55
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@5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
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– Arthur
Jan 8 at 9:56
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@Arthur " That's what I asked earlier"... I don't see your question anywhere...
$endgroup$
– 5xum
Jan 8 at 9:58
|
show 1 more comment
$begingroup$
The question I met is to show that if $z=cos (theta)+isin(theta)$ with $i=sqrt{-1}$,then $ Re(frac{z-1}{z+1})=0$
In the normal way, we found that: $$frac{z-1}{z+1}=frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}\
=frac{bigl(cos(theta)-1+isin(theta)bigl)bigl(cos(theta)+1-isin(theta)bigl)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{cos^2(theta)+cos(theta)-cos(theta)-1+sin^2(theta)+2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}$$
So $Re(frac{z-1}{z+1})=0$
If we do it in another way:
$$
frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}=frac{-2sin^2(frac{theta}{2})+2icos(frac{theta}{2})sin(frac{theta}{2})}{2cos^2(frac{theta}{2})+2isin(frac{theta}{2})cos(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{sin(frac{theta}{2})-icos(frac{theta}{2})}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)
$$
So the real part of it will be $-tan(frac{theta}{2})cos(-theta-frac{pi}{2})$ which is not $0$.
Which step I made mistake or they are equivalent?
complex-numbers
asked Jan 8 at 9:38
yuanming luoyuanming luo
10211
$endgroup$
The question I met is to show that if $z=cos (theta)+isin(theta)$ with $i=sqrt{-1}$,then $ Re(frac{z-1}{z+1})=0$
In the normal way, we found that: $$frac{z-1}{z+1}=frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}\
=frac{bigl(cos(theta)-1+isin(theta)bigl)bigl(cos(theta)+1-isin(theta)bigl)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{cos^2(theta)+cos(theta)-cos(theta)-1+sin^2(theta)+2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}$$
So $Re(frac{z-1}{z+1})=0$
If we do it in another way:
$$
frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}=frac{-2sin^2(frac{theta}{2})+2icos(frac{theta}{2})sin(frac{theta}{2})}{2cos^2(frac{theta}{2})+2isin(frac{theta}{2})cos(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{sin(frac{theta}{2})-icos(frac{theta}{2})}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)
$$
So the real part of it will be $-tan(frac{theta}{2})cos(-theta-frac{pi}{2})$ which is not $0$.
Which step I made mistake or they are equivalent?
complex-numbers
complex-numbers
asked Jan 8 at 9:38
yuanming luoyuanming luo
10211
asked Jan 8 at 9:38
yuanming luoyuanming luo
10211
edited Jan 8 at 9:47
yuanming luo
asked Jan 8 at 9:38
yuanming luoyuanming luo
10211
asked Jan 8 at 9:38
yuanming luoyuanming luo
10211
asked Jan 8 at 9:38
yuanming luoyuanming luo
10211
10211
$begingroup$
There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
$endgroup$
– Yves Daoust
Jan 8 at 9:44
$begingroup$
Demoivre's theory, the arguments of two complex number are added.
$endgroup$
– yuanming luo
Jan 8 at 9:47
$begingroup$
How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
$endgroup$
– 5xum
Jan 8 at 9:55
$begingroup$
@5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
$endgroup$
– Arthur
Jan 8 at 9:56
$begingroup$
@Arthur " That's what I asked earlier"... I don't see your question anywhere...
$endgroup$
– 5xum
Jan 8 at 9:58
|
show 1 more comment
$begingroup$
There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
$endgroup$
– Yves Daoust
Jan 8 at 9:44
$begingroup$
Demoivre's theory, the arguments of two complex number are added.
$endgroup$
– yuanming luo
Jan 8 at 9:47
$begingroup$
How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
$endgroup$
– 5xum
Jan 8 at 9:55
$begingroup$
@5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
$endgroup$
– Arthur
Jan 8 at 9:56
$begingroup$
@Arthur " That's what I asked earlier"... I don't see your question anywhere...
$endgroup$
– 5xum
Jan 8 at 9:58
$begingroup$
There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
$endgroup$
– Yves Daoust
Jan 8 at 9:44
$begingroup$
There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
$endgroup$
– Yves Daoust
Jan 8 at 9:44
$begingroup$
Demoivre's theory, the arguments of two complex number are added.
$endgroup$
– yuanming luo
Jan 8 at 9:47
$begingroup$
Demoivre's theory, the arguments of two complex number are added.
$endgroup$
– yuanming luo
Jan 8 at 9:47
$begingroup$
How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
$endgroup$
– 5xum
Jan 8 at 9:55
$begingroup$
How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
$endgroup$
– 5xum
Jan 8 at 9:55
$begingroup$
@5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
$endgroup$
– Arthur
Jan 8 at 9:56
$begingroup$
@5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
$endgroup$
– Arthur
Jan 8 at 9:56
$begingroup$
@Arthur " That's what I asked earlier"... I don't see your question anywhere...
$endgroup$
– 5xum
Jan 8 at 9:58
$begingroup$
@Arthur " That's what I asked earlier"... I don't see your question anywhere...
$endgroup$
– 5xum
Jan 8 at 9:58
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
You should be more careful when applying transformations of sines into cosines.
It's much easier than that:
$$
sinfrac{theta}{2}-icosfrac{theta}{2}=
ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
$$
You could write this in trigonometric form
$$
=left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
=cosleft(frac{theta}{2}+frac{pi}{2}right)+
isinleft(frac{theta}{2}+frac{pi}{2}right)
$$
and now you can go and chase for your mistake.
Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
$$
bar{w}=frac{bar{z}-1}{bar{z}+1}
$$
but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
$$
bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
$$
From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.
If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
$$
w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
=frac{u-bar{u}}{u+bar{u}}=
frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
$$
answered Jan 8 at 10:05
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
We have
$$
sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
= -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
= -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
$$
so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.
answered Jan 8 at 9:56
ArthurArthur
123k7122211
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You should be more careful when applying transformations of sines into cosines.
It's much easier than that:
$$
sinfrac{theta}{2}-icosfrac{theta}{2}=
ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
$$
You could write this in trigonometric form
$$
=left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
=cosleft(frac{theta}{2}+frac{pi}{2}right)+
isinleft(frac{theta}{2}+frac{pi}{2}right)
$$
and now you can go and chase for your mistake.
Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
$$
bar{w}=frac{bar{z}-1}{bar{z}+1}
$$
but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
$$
bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
$$
From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.
If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
$$
w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
=frac{u-bar{u}}{u+bar{u}}=
frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
$$
answered Jan 8 at 10:05
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
You should be more careful when applying transformations of sines into cosines.
It's much easier than that:
$$
sinfrac{theta}{2}-icosfrac{theta}{2}=
ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
$$
You could write this in trigonometric form
$$
=left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
=cosleft(frac{theta}{2}+frac{pi}{2}right)+
isinleft(frac{theta}{2}+frac{pi}{2}right)
$$
and now you can go and chase for your mistake.
Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
$$
bar{w}=frac{bar{z}-1}{bar{z}+1}
$$
but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
$$
bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
$$
From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.
If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
$$
w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
=frac{u-bar{u}}{u+bar{u}}=
frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
$$
answered Jan 8 at 10:05
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
You should be more careful when applying transformations of sines into cosines.
It's much easier than that:
$$
sinfrac{theta}{2}-icosfrac{theta}{2}=
ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
$$
You could write this in trigonometric form
$$
=left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
=cosleft(frac{theta}{2}+frac{pi}{2}right)+
isinleft(frac{theta}{2}+frac{pi}{2}right)
$$
and now you can go and chase for your mistake.
Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
$$
bar{w}=frac{bar{z}-1}{bar{z}+1}
$$
but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
$$
bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
$$
From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.
If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
$$
w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
=frac{u-bar{u}}{u+bar{u}}=
frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
$$
answered Jan 8 at 10:05
egregegreg
186k1486208
$endgroup$
You should be more careful when applying transformations of sines into cosines.
It's much easier than that:
$$
sinfrac{theta}{2}-icosfrac{theta}{2}=
ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
$$
You could write this in trigonometric form
$$
=left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
=cosleft(frac{theta}{2}+frac{pi}{2}right)+
isinleft(frac{theta}{2}+frac{pi}{2}right)
$$
and now you can go and chase for your mistake.
Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
$$
bar{w}=frac{bar{z}-1}{bar{z}+1}
$$
but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
$$
bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
$$
From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.
If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
$$
w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
=frac{u-bar{u}}{u+bar{u}}=
frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
$$
answered Jan 8 at 10:05
egregegreg
186k1486208
answered Jan 8 at 10:05
egregegreg
186k1486208
answered Jan 8 at 10:05
egregegreg
186k1486208
answered Jan 8 at 10:05
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
$begingroup$
We have
$$
sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
= -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
= -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
$$
so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.
answered Jan 8 at 9:56
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
We have
$$
sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
= -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
= -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
$$
so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.
answered Jan 8 at 9:56
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
We have
$$
sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
= -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
= -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
$$
so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.
answered Jan 8 at 9:56
ArthurArthur
123k7122211
$endgroup$
We have
$$
sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
= -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
= -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
$$
so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.
answered Jan 8 at 9:56
ArthurArthur
123k7122211
answered Jan 8 at 9:56
ArthurArthur
123k7122211
answered Jan 8 at 9:56
ArthurArthur
123k7122211
answered Jan 8 at 9:56
ArthurArthur
123k7122211
123k7122211
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$begingroup$
There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
$endgroup$
– Yves Daoust
Jan 8 at 9:44
$begingroup$
Demoivre's theory, the arguments of two complex number are added.
$endgroup$
– yuanming luo
Jan 8 at 9:47
$begingroup$
How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
$endgroup$
– 5xum
Jan 8 at 9:55
$begingroup$
@5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
$endgroup$
– Arthur
Jan 8 at 9:56
$begingroup$
@Arthur " That's what I asked earlier"... I don't see your question anywhere...
$endgroup$
– 5xum
Jan 8 at 9:58