Solving the equation $\sinh(z) +3\cosh(z) = 1$

Question

I’m trying to find the complex solutions for the equation

$$\sinh(z) + 3\cosh(z) = 1$$

Using Wolfram Alpha I know that there exists a solution, but I don’t know how to get there on my own…

Many thanks for any help,

Felix

Answer 1

\begin{align} \sinh z + 3\cosh z &= \frac{1}{2}\left(e^{z}-e^{-z}+3e^{z}+3e^{-z}\right)\\ &=\frac{1}{2}\left(4e^{z}+2e^{-z}\right)\\ &=2e^{z}+e^{-z} \end{align}

Therefore, we are looking for a $z$ such that

$$2e^{z}+e^{-z} = 1$$

This is equivalent to

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

which we can rearrange to

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

Substituting $u = e^{z}$, we have

$$2u^2 - u + 1 = 0$$

which has the solutions

$$u_{1/2} = \frac{1 \pm\sqrt{-7}}{4} = \frac{1\pm i\sqrt{7}}{4}$$

Now recall that $u = e^{z}$, we have the final solutions

$$z_{1/2} = \log\left({\frac{1\pm i\sqrt{7}}{4}}\right)$$

Answer 2

Hint.

$$ \frac{e^z-e^{-z}}{2}+3\frac{e^z+e^{-z}}{2} = 1 \Rightarrow 2e^z+e^{-z} = 1 $$

etc.

Answer 3

The right answer was $z = \ln(1/4 + i\cdot\sqrt{7}/4)$. This solution plus $2\pi n$ is also solution of the equation above.