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If $z=x+iy$ is the answer of the equation $\cos(z)=-3$. What is the value of $x$?

$1)x=2k\pi-\frac{\pi}2 , k\in \mathbb{Z}$

$2)x=2k\pi+\frac{\pi}2 , k\in \mathbb{Z}$

$3)x=k\pi+\frac{\pi}2 , k\in \mathbb{Z}$

$4)x=k\pi-\frac{\pi}2 , k\in \mathbb{Z}$

I saw a similar question here, and I tried to solve it but I got stock at a point. Here is my try:

$$\cos z=\frac{e^{iz}+e^{-iz}}{2}=-3\Rightarrow e^{iz}+e^{-iz}+6=0$$ After using the substitution $e^{iz}=t$ we get $t^2+6t+1=0$. Hence $e^{i(z+2k\pi)}= -3\pm\sqrt8$ and

$$i(z+2k\pi)=\ln(-3\pm2\sqrt2)$$ From here I don't know how to continue.

Etemon
  • 6,437
  • We are all agreed that the correct answer is None of the Above (specifically, $2k\pi+\pi$). Are you sure you copied the question correctly? – TonyK May 15 '22 at 16:26
  • @TonyK Yes I just double checked it and I copied correctly. – Etemon May 15 '22 at 16:29

2 Answers2

1

$$e^{iz} = e^{i(x+iy)}= -3\pm\sqrt8 \implies \underbrace{e^{-y}}_{>0}\ e^{ix}= -3\pm\sqrt8<0$$

$$ \implies e^{ix}\in \mathbb{R}_{<0} \implies x=(2k+1)\pi,\ k\in\mathbb{Z}.$$

Adam Rubinson
  • 20,052
1

\begin{align} z &= 2\pi k \pm \cos^{-1}(-3), \; k \in \mathbb{Z}\\ &= 2\pi k \pm \frac{\pi}{2} \pm \sin^{-1}(3)\\ & = 2\pi k \pm \frac{\pi}{2} \pm i\ln\left(\sqrt{1 - 3^2} - 3i\right)\\ & = 2\pi k \pm \frac{\pi}{2} \pm i\ln\left(i\left(2\sqrt{2} - 3\right)\right)\\ & = 2\pi k \pm \frac{\pi}{2} \pm i\ln(i) \pm i\ln\left(2\sqrt{2} - 3\right)\\ & = 2\pi k \pm \frac{\pi}{2} \mp \frac{\pi}{2} \pm i\ln\left(2\sqrt{2} - 3\right)\\ & = 2\pi k \pm i\ln\left(2\sqrt{2} - 3\right)\\ & = 2\pi k \mp \pi \pm i\ln\left(3 - 2\sqrt{2}\right) \end{align}

$$\therefore \boxed{\Re(z) = (2n + 1)\pi, \; n \in \mathbb{Z}}$$

Reference: Inverse trigonometric functions: Logarithmic forms