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Given: $$\sin(x+iy)=\cos\theta+i\sin\theta$$ To prove: $$x=\arccos (\sqrt{\sin\theta})$$ How I tried: $$\begin{align*} \sin x \cosh y &= \cos\theta \\ \cos x \sinh y &= \sin\theta \end{align*}$$ Then tried to use logarithm of hyperbolic complex number.

Also various trignometric form manipulation but I can't get the answer.

alexwlchan
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Ashish
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  • What exactly is y ? – Lucian Dec 01 '14 at 19:06
  • x and y both are real quantities. The question was asked in university exam late back in 2002, and not much was given in the question. I think question could also be written as x+iy=arcsin(e^i∅) then finding x and y – Ashish Dec 02 '14 at 01:04

2 Answers2

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given:

$sin(x+iy)=e^{i\theta}$

$sin(x)*cosh(y)+icos(x)*sinh(y)=cos(\theta)+isin(\theta)$

comparing the the real and imaginary part

$\implies sin(x)*cosh(y)=cos(\theta)$

&

$\implies cos(x)*sinh(y)=sin(\theta)$

consider hyperbolic form of $sin^2y +cos^2y=1 $

$\cosh^2y-sinh^2y=1$

put values of $sinh(y) \& cosh(y)$ in above equation

$ \frac {\cos^2\theta}{ \sin^2x} $-$\frac{sin^2\theta}{cos^2x}$=1

cross muliplying and arranging

$cos^2\theta*cos^2x-sin^2\theta*sin^2x=sin^2x*cos^2x$

$(1-sin^2\theta)*cos^2x-sin^2\theta*sin^2x=cos^2x*(1-cos^2x)$

$cos^2x-sin^2\theta(cos^2x+sin^2x)=cos^2x-cos^2x*cos^2x $

$-sin^2\theta=-cos^4x$

taking twice squre root

$cosx=\sqrt\sin\theta$

Ashish
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  • This does not make much sense. If you start with an arbitrary complex number $x+iy$, then $θ$ is also complex. You state that $θ$ is real, but you did not do so when you posted the problem!! Furthermore, if $sin(x+iy)$ is on the unit circle, we can immediately establish that $sinh(y)= +/- cos(x)$. No need for your long derivation. The final answer is also incorrect. Actually $sin(θ)= +/- (cos(x))^2$, with the condition that $cos(θ)$ has the same sign as $sin(x)$. You should be more careful with your minus signs. – M. Wind Dec 03 '14 at 16:49
  • Your derivation is essentially reasonable. Three aspects: You should clearly state right at the beginning the domain of definition, presumably $x,y,\theta\in \mathbb{R}$. You divide by $\sin ^2x$ and $\cos^2x$, so you have to explicitely analyse the situation when the denominator is zero. Taking square root twice implies you get more than one solution. Therefore you should carefully consider all solutions in order to completely elaborate the claim. Nevertheless +1 for your reasonable approach. Regards, – Markus Scheuer Dec 03 '14 at 23:25
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We start off with two real numbers $x$ and $y$, and consider the complex function $\sin(x+iy)$. As has been pointed out by others in this thread, this is a complex number that can written as $z = \sin(x + iy) = \sin(x)\cosh(y)+i\cos(x)\sinh(y)$.

We are then asked to express this result in terms of a new variable $\theta$ by means of the relation $z = e^{i\theta} = \cos(\theta) + i\sin(\theta)$. In general this is only possible if $\theta$ itself is a complex number, and equally so for the functions in which it appears. For example $\sin(\theta)$ can be written as $\sin(\theta) = \frac{(z - 1/z)}{2i}$.

However, the OP has now stated that $\theta$ is a real. This implies that $z$ lies on the unit circle. Using the fact that the norm of $z$ is equal to one, we see $x$ and $y$ are related by: $\sinh(y) = \pm\cos(x)$. Note that this equation can only have a solution when $\sinh(y)$ is in the interval $(-1, 1)$, hence values of $y$ are restricted to the interval $(-0.881377, +0.881377)$.

The two branches of the solution are described by the following equations.

A) $y = +arc\sinh(\cos(x))$. Also $\sin(\theta) = \cos^2(x)$ and $\cos(\theta) = \sin(x)\sqrt{1+cos^2(x)}$

B) $y = -arc\sinh(\cos(x))$. Also $\sin(\theta) = -\cos^2(x)$ and $\cos(\theta) = \sin(x)\sqrt{1+cos^2(x)}$

We see that in both branches $y$ and $\theta$ are periodic functions of $x$. If we combine the two branches in a single plot, $y$ and $\theta$ oscillate perfectly in phase, each having period $\pi$. The amplitude of $y$ is $arc\sinh(1) = 0.881377$ and the amplitude of $\theta$ is $\pi/2$. Note however that this interpretation does not apply to the separate branches, because then the sign of $\theta$ remains the same.

Leyla Alkan
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M. Wind
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  • it is solvable and $e^{i\theta}=cos\theta+i sin\theta $ is euler's form and does not mean $\theta$ is complex number is is an angle – Ashish Dec 03 '14 at 10:09
  • Thanks for your efforts, I am beginner in complex number I know there there was some detail missing in my question but I have not down voted you. – Ashish Dec 04 '14 at 13:33