Solve $2\cosh z + \sinh z = i$ for $z$ The equation is $$2\cosh z + \sinh z = i$$
I used the following formulas:
$$\cosh z = \frac{e^z + e^{-z}}{2},
  \sinh z = \frac{e^z - e^{-z}}{2}$$
to reduce this equation to $$3e^z - e^{-z} = 2i$$ but I am not sure now to proceed now. I had idea about replacing $e^z$ with $x$, having $$3e^z -2i - e^{-z} = 0 \implies 3x^2 -2ix - 1 = 0$$
but I am to sure I can do this, and resulting $x$ is not looking good. My teacher said this equation can be reduced to quadratic, but he didn't explain it in detail. How do I solve this?
 A: $$2\cosh(z)+ \sinh(z) = i$$
$$e^z + e^{-z} + \frac{e^z - e^{-z}}{2} = i$$
$$2e^z + 2e^{-z} + e^z - e^{-z} = 2i$$
$$3e^z+e^{-z}= 2i$$
$$3e^z+(e^{z})^{-1}= 2i$$
Let $e^z = x$.
$$3x+x^{-1} = 2i$$ 
$$3x^2+1 = 2ix$$ 
$$3x^2-2ix+1 = 0$$ 
We solve equations of the form $ax^2+bx+c=0$ with 
$$x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
By plugging in $a=3$, $b=-2i$ and $c=1$:
$$x_{1,2} = \frac{2i \pm \sqrt{-4-4\cdot 3 \cdot 1}}{6}$$
$$ = \frac{2i \pm \sqrt{-4-12}}{6} = \frac{2i \pm \sqrt{-16}}{6}= \frac{2i \pm 4i}{6} = \frac{i}{3} \pm \frac{2i}{3}$$
The two roots are therefore
$$x_1 = i, \quad x_2 = -\frac{i}{3}$$
Since we required $e^z = x$, $z = \text{Ln}(x)$ ($\text{Ln}$ being the complex-valued logarithm). We are lazy and throw both values in Wolfram Alpha and get
$$z_1 = \ln(i) = \frac{i\pi}{2},\quad z_2 = \ln\left(-\frac{i}{3}\right) = -\ln(3) - \frac{i\pi}{2}$$
And now we have $z_1$ and $z_2$, which are the solutions to the initial equation. Notice that the solution to that complex logarithm has arbitrary multiples of $2\pi i$ in it (i.e. it is multivalued). 
A: I think you are going fine. Use Quadratic formula to solve for $x$ and then take log
