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Find all number $z$ such that $\cos(z) = i\sin(z)$.

Options : $$(a) \ in, \ n\in \mathbb{Z} \\ (b) \ inπ +1, \ n\in \mathbb{Z} \\ (c) \ \frac{inπ}{2}, \ n\in \mathbb{Z} \\ (d) \ \text{no solutions exist}.$$

My try : Let $z=x+iy$.

Then, $\cos(z) = \frac{\exp(iz)+exp(-iz)}{2}$ and $\sin(z) = \frac{\exp(iz)-exp(-iz)}{2}$.

$$\therefore \cos(z) = i\sin(z) \\ \implies \frac{\exp(iz)+exp(-iz)}{2} = i \frac{\exp(iz)-exp(-iz)}{2} \\ \implies \exp(iz)\big(1-i\big) = \exp(-iz)\big(-1-i) \\ \implies \frac{\exp(iz)}{exp(-iz)} = - \frac{1+i}{1-i} = -i \\ \implies \exp(i2z) = -i $$

Now, $$i2z = -2y + i 2x$$ $$\therefore \exp(i2z) = -i \\ \implies e^{-2y}\Big(\cos(2x) + i \sin(2x)\Big) = -i$$

Therefore, $$e^{-2y}\cos(2x) = 0, e^{-2y}\sin(2x) = -1 $$ We have, $$ e^{-2y} = 1 \\ \implies y = 0$$ and $$ \cos(2x) = 0, \sin(2x) = -1 \\ \implies 2x = \Big( 2nπ + \frac{3π}{2}\Big) \\ \implies x = \Big(4nπ +3\Big) \frac{π}{4} $$

Thus the solution are the numbers, $$z = \Big(4nπ +3\Big) \frac{π}{4} , \ \ n\in \mathbb{Z}$$.

But when I'm plugging this in the given equation,. I'm getting nonsense.

Can anyone tell me where I'm going wrong?

Itachi
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    Note that $sin(z)=\frac{\exp(iz)-\exp(-iz)}{2i}$ and not $sin(z)=\frac{\exp(iz)-\exp(-iz)}{2}$. – YumekuiMath Dec 20 '21 at 08:20
  • Thank you. I totally missed it. – Itachi Dec 20 '21 at 08:24
  • @YumekuiMath but doing that gives me no solution. As $\exp(-iz) \ne 0$ for any $z$. Am I correct? – Itachi Dec 20 '21 at 08:35
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    Yes, that is correct. – YumekuiMath Dec 20 '21 at 08:43
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    @Martin I already answered the question in the comments. – Itachi Dec 20 '21 at 09:15
  • @Itachi: Comments can be deleted at any time and their purpose is to clarify a problem, not to answer a question. If you don't need an answer then you can delete the question. Or we close it as a duplicate, because an almost identical question has answers. – Martin R Dec 20 '21 at 09:32
  • @Martin : I'd have deleted the question but as it's already answered so it's showing that I need to vote to delete the question. Willi you then help me get some votes? – Itachi Jan 14 '22 at 12:50

2 Answers2

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Essentially by definition, $$ \cos z+i\sin z=e^{iz} \tag{1} $$ Changing $z$ into $-z$, we get $$ \cos z-i\sin z=e^{-iz} \tag{2} $$ and this is nonzero for every $z\in\mathbb{C}$.

Note that from $(1)$ and $(2)$ we obtain $$ \sin z=\dfrac{e^{iz}-e^{-iz}}{2i} $$ that you want to compare with what you wrote.

egreg
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From $\cos(z)=\dfrac{e^{iz}+e^{-iz}}2$ and $\sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}$, your equation becomes:

$$e^{iz}+e^{-iz}=e^{iz}-e^{-iz}$$

Hence

$$e^{-iz}=0$$

Conclusion?

(if in doubt about the conclusion, note that $e^ze^{-z}=1$ for all $z$)

Jean-Claude Arbaut
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