Find all the complex roots of $\cosh(z)=\frac{1}{2}$

Question

Find all the complex roots of $\cosh(z)=\frac{1}{2}$

I have so far used the formula $\cosh(z)=\frac{e^z+e^{-z}}{2}$ to get:

\begin{align} & \frac{e^z+e^{-z}}{2}=\frac{1}{2} \\ \implies & e^z+e^{-z}=1 \end{align}

Now, I'm not sure where to go.

Answer 1

You can rewrite the expression with a common denominator, since $e^{-z} = \dfrac{1}{e^{z}}$:

$$\dfrac{(e^{z})^{2}}{e^{z}}+\dfrac{1}{e^{z}}=1$$

Multiply by $e^{z}$ on both sides:

$$(e^{z})^{2} + 1 = e^{z}$$

Then subtract both sides by $e^{z}$:

$$(e^{z})^{2} - e^{z} + 1 = 0$$

Solve via the quadratic formula:

$$e^{z} = \dfrac{1\pm\sqrt{-3}}{2}$$ $$z = \ln\left(\dfrac{1\pm i\sqrt{3}}{2}\right)$$

And taking the natural log of a complex number $z$ is $\ln{|z|} + i\arg{z}$, giving:

$$z = \pm\;i\dfrac{\pi}{3}\pm2k\pi$$

Answer 2

The method you are following is pretty standard, here are (the beginnings of) two others.

• $\cosh z=\frac12$, so $\sinh^2z=\cosh^2z-1=-\frac34$, so $\sinh z=\pm i\frac{\sqrt3}2$, so $$e^z=\cosh z+\sinh z=\frac{1\pm i\sqrt3}2\ .$$
• $\cos(iz)=\cosh z=\frac12$, so $$iz=\pm\frac\pi3\pm 2k\pi\ .$$

I'll also continue your argument a bit - wasn't going to as it was in someone else's answer, but that answer has now been deleted (don't know why, there was nothing wrong with it).

• Multiply both sides of $e^z+e^{-z}=1$ by $e^z$ and rearrange to get $$(e^z)^2-(e^z)+1=0$$ then solve the quadratic.

I'll leave you to finish the details.

Answer 3

$\displaystyle e^z+e^{-z}=1 \tag 1$

$\displaystyle z=x+iy \tag 2$

From $1$ and $2$ you can say that:

$\displaystyle e^z+e^{-z}=e^{x+iy}+e^{-x-iy}=e^{x}(e^{iy})+e^{-x}(e^{-iy})=e^x(cos(y)+isin(y))+e^{-x}(cos(y)-isin(y))=\cos(y)(e^x+e^{-x})+i\sin(y)(e^x-e^{-x})$

$\displaystyle=2\cos(y)\cosh(x)+2i\sin(y)\sinh(x)=1+0i$

$\displaystyle \to \sin(y)\sinh(x)=0$ and $\displaystyle \cos(y)\cosh(x)=\dfrac12$

$\sinh(x)=0, \cos(y)=\dfrac12$, So:

$\displaystyle x=0$ and $\displaystyle y=2k\pi\pm \pi/3,$ or simply: $\displaystyle z=2k\pi\pm \pi/3$