# Math Help - Complex equation

1. ## Complex equation

Hi everybody

I need some help to solve this complex equation:

(√3 + i − ez) (ez-1) = 0

that has a modulus smaller than 3π.

I'm thinking about using zero product property:

ez - 1 = 0

z = ln(1)

z = 0

but when it comes to the first parenthesis I don't know how to proceed.

Thank you.

2. ## Re: Complex equation

Originally Posted by Tala
Hi everybody

I need some help to solve this complex equation:

(√3 + i − ez) (ez-1) = 0

that has a modulus smaller than 3π.

I'm thinking about using zero product property:

ez - 1 = 0

z = ln(1)

z = 0

but when it comes to the first parenthesis I don't know how to proceed.

Thank you.
If we remember that \displaystyle \begin{align*} z = x + i\,y \end{align*} with \displaystyle \begin{align*} x, y \in \mathbf{R} \end{align*}, setting the first factor equal to 0 gives

\displaystyle \begin{align*} \sqrt{3} + i - e^z &= 0 \\ \sqrt{3} + i &= e^z \\ 2e^{\frac{\pi}{6} i} &= e^{x + i\,y} \\ 2e^{\frac{\pi}{6} i} &= e^x e^{i\,y} \\ e^x = 2 \textrm{ and } e^{\frac{\pi}{6} i } &= e^{i\,y} \\ x = \ln{2} \textrm{ and } y &= \frac{\pi}{6} \end{align*}

Therefore \displaystyle \begin{align*} z = \ln{2} + \frac{\pi}{6} i \end{align*}

3. ## Re: Complex equation

This might be a stupid question but I can't see how the argument is π/6 ?

4. ## Re: Complex equation

I get it know:

arccos(√(3)/2) right ?

5. ## Re: Complex equation

Originally Posted by Tala
This might be a stupid question but I can't see how the argument is π/6 ?
If the complex number is in the first quadrant, which it is, \displaystyle \begin{align*} \textrm{arg}\,(z) = \arctan{\left[ \frac{\textrm{Im}\,(z)}{\textrm{Re}\,(z)} \right]} \end{align*}.

6. ## Re: Complex equation

Thank you so much for the quick and brilliant answers.
I really appreciate it.
Tak (it means thank you in danish).

7. ## Re: Complex equation

Originally Posted by Prove It
Therefore \displaystyle \begin{align*} z = \ln{2} + \frac{\pi}{6} i \end{align*}
It is worth noting that for each $k\in\mathbb{Z},~\ln (2) + i\left( {\frac{\pi }{6} + 2k\pi } \right)$ is also a solution.