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The equation:
$$e^{iz} = 4\cos(z) + 3$$ This is what I have tried so far: $$e^{iz} = 4\left(\frac{e^{iz}+e^{-iz}}{2}\right)+3$$ $$e^{iz} = 2e^{iz}+2^{-iz}+3$$ $$e^{2iz} = 2e^{2iz}+2+3e^{iz}$$ $$e^{2iz}+3e^{iz}+2=0$$ Now, I substitute $e^{iz}$ with $x$, resulting in $x=-1$ or $x=-2$. This leaves me at $e^{iz}=-1$ or $e^{iz}=-2$
I'm not sure how to continue from here. I can get rid of the $e$ by taking the natural logarithm, but that makes further calculations very difficult.

GNUSupporter 8964民主女神 地下教會
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B.E
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    Take logarithms, as you said. For example, $\ln(-1)=\ln(e^{\pi i + 2k\pi i})=\pi i + 2k\pi i$ for $k\in\mathbb{Z}$. Therefore $e^{iz}=-1$ gives you the solutions $z=\pi+2k\pi$ for $k\in\mathbb{Z}$. The key step is the first: writing $-1$ in its polar form $e^{i\pi}$. The polar form of $-2$ is $2e^{i\pi}$. –  Mar 14 '18 at 19:40
  • Wow, that really helps. For $z=-2$ I end up with $z = ln(2)\pi + 2ki\pi$, is that correct? – B.E Mar 14 '18 at 19:55
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    Yes, almost: $z=2k\pi+\pi-i\ln2$. – Berci Mar 14 '18 at 20:24
  • Do you get to $-i ln(2)$ by multiplying $\frac{ln(2)}{i}$ by $\frac{-i}{-i}$? – B.E Mar 17 '18 at 10:00
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    Yes, that shows $\frac{\ln2}i=-i\ln2$. – Berci Mar 17 '18 at 10:19

1 Answers1

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We use the following well-known properties of the complex exponential function: $\forall z \in \mathbb{C}, \forall z' \in \mathbb{C}:$

  • $e^{z} e^{z'}=e^{z+z'}$;
  • $e^z=1\iff\exists k\in \mathbb{Z}, z=2ik\pi$.

From these properties, it follows that : $\forall z \in \mathbb{C}, e^z\neq0$; $\forall z \in \mathbb{C}, \forall z' \in \mathbb{C}, e^z=e^{z'}\iff \exists k\in \mathbb{Z},z=z'+2ik\pi$.

So, $e^{iz}=-1=e^{i\pi}\iff \exists k\in \mathbb{Z},iz=i\pi+2ik\pi\iff \exists k\in \mathbb{Z},z=\pi+2k\pi$ $e^{iz}=-2=e^{i\pi}e^{log2}=e^{i\pi+log2}\iff \exists k\in \mathbb{Z},iz=i\pi+log2+2ik\pi\iff \exists k\in \mathbb{Z},z=\pi+2k\pi-ilog2$

Stéphane Jaouen
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