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I am to spot the issue with the following problem: Since $e^{iz} = cos (z)+ i \cdot sin(z)$ follows that $$\vert e^{iz}\vert = \sqrt{cos^2(z)+sin^2(z)} = 1$$ Hence, if $z=i$, we can derive that $\vert e^{iz}\vert = 1/e$ and thus $e=1$.
I have to find the mistake.

I am (merely) guessing that the following part $\vert e^{iz}\vert = \sqrt{cos^2(z)+sin^2(z)}$ is incorrect, however, I am having troubles proving this.
My current approach is as follows:

$$\vert e^{iz}\vert = \vert \cos(z)+i\sin(z)\vert\\ = \sqrt{(\cos(z)+i\sin(z))(\overline{\cos(z)+i \sin(z)})}$$ if $z=x+iy$: $$ = \sqrt{(\cos(x+iy)+i\sin(x+iy))(cos(x-iy)-i\sin(x-iy))} \\ = \sqrt{\cos(x+iy)\cos (x-iy)-\cos(x+iy)i\sin((x-iy)+i\sin(x+iy)\cos(x-iy)-\sin(x+iy)\sin(x-iy)}$$

And since $\sin(x+yi) = \sin x \cos (yi) + \cos(x) \sin(yi)$ and $\cos(x+yi) = \cos x \cos (yi) - \sin(x) \sin(yi)$ I could continue calculating but this does not seem to be the correct way. I would be very grateful for any hints or ideas. Thanks.

primumxyz
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    Well, the modulus of a complex number $z=a+ib$ is $\sqrt{a^2+b^2}$, but only if $a$ and $b$ are real. That is, $a$ is the real part of $z$ and $b$ is the imaginary part. What is always true: $|z|=\sqrt{\mathrm{Re}(z)^2+\mathrm{Im}(z)^2}$. – Jean-Claude Arbaut Dec 13 '22 at 21:37

1 Answers1

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Let $z=x+iy$. Then

$$\begin{align*} e^{iz} &= e^{ix-y} \\ &= e^{-y} (\cos(x) + i \sin(x)) \\[1ex] \implies |e^{iz}| &= \sqrt{\left(e^{-y}\cos(x)\right)^2 + \left(e^{-y}\sin(x) \right)^2} \\ &= e^{-y} \end{align*}$$

but $\cos^2(z)+\sin^2(z)=1$ for all $z\in\Bbb C$, so the equation

$$\left|e^{iz}\right| = \sqrt{\cos^2(z) + \sin^2(z)}$$

is true only when $y=0$, i.e. $z\in\Bbb R$.

user170231
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