Find every $z$s that fit $\cos(z) = -2$

Question

I couldn't find any, I tried to write $\cos(z)$ as $\cos(x)\cos(iy)-\sin(x)\sin(iy)$ which then gave me

$\cos(x)\cosh(y) - i\sin(x)\sinh(y) = -2$

$\sin(x)=0$ so that imaginary part become $0$

now we have to find $\cosh(y) = -2$ which is not true for no $y$.

is it right or i made a mistake in my substitutions?

Answer 1

Hint: start with $$ \cos z=\frac{e^{iz}+e^{-iz}}{2} $$ and obtain a quadratic equation for $e^{iz}$.

Answer 2

Be: $$ \cos z=\frac{e^{iz}+e^{-iz}}{2}=-2 $$ Then: \begin{eqnarray} \frac{e^{iz}+e^{-iz}}{2} &=& -2 \\ e^{iz}+e^{-iz} &=& -4 \\ e^{2iz} + 1 &=& -4e^{iz} \\ \left(e^{iz}\right)^2 + 4\left(e^{iz}\right) + 1 &=& 0 \\ e^{iz} &=& \frac{-4 \pm \sqrt{16-4}}{2} \\ e^{iz} &=& -2 \pm \sqrt{3} \\ \end{eqnarray} Then since $e^{iz}=\cos(z)+i\sin(z)$, $e^{iz}=e^{i(z+2k\pi)}$ with $k\in\mathbb{Z}$: \begin{eqnarray} e^{i(z+2k_1\pi)} &=& -2 \pm \sqrt{3} \\ i(z +2k_1\pi) &=& \ln\left(-2 \pm \sqrt{3}\right) \\ z &=& 2k_1\pi - i\ln\left(-2 \pm \sqrt{3}\right) \\ \end{eqnarray} There are two set solution to differents solutions: \begin{eqnarray} z_{+} &=& 2k_1\pi - i\ln\left(-2 + \sqrt{3}\right) \\ z_{-} &=& 2k_1\pi - i\ln\left(-2 - \sqrt{3}\right) \\ \end{eqnarray} For $z_{-}$: \begin{eqnarray} z_{-} &=& 2k_1\pi - i\ln\left(\left(2 + \sqrt{3}\right)e^{-i(1+2k_2)\pi}\right) \\ z_{-} &=& 2k_1\pi - i\left[\ln\left(2 + \sqrt{3}\right)-i(1+2k_2)\pi\right] \\ z_{-} &=& 2k_1\pi - (1+2k_2)\pi - i\ln\left(2 + \sqrt{3}\right) \\ z_{-} &=& (1+2k)\pi - i\ln\left(2 + \sqrt{3}\right) \\ \end{eqnarray} Then, with $k\in\mathbb{Z}$ the solutions are: \begin{eqnarray} z_{+} &=& 2k\pi - i\ln\left(-2 + \sqrt{3}\right) \\ z_{-} &=& (1+2k)\pi - i\ln\left(2 + \sqrt{3}\right) \\ \end{eqnarray}

Answer 3

I should have used $x=(2k+1)\pi$ for the $x$ and then had the

$\cos( (2k+1)\pi ) \cosh(y) = -2$

which made it $\cosh(y) = 2$