Complex equation

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September 17th 2012, 07:59 AM
Tala

Complex equation

Hi everybody

I need some help to solve this complex equation:

(√3 + i − e^z) (e^z-1) = 0

that has a modulus smaller than 3π.

I'm thinking about using zero product property:

e^z - 1 = 0

z = ln(1)

z = 0

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

Thank you.
September 17th 2012, 08:07 AM
Prove It

Re: Complex equation

Quote:

Originally Posted by Tala

Hi everybody

I need some help to solve this complex equation:

(√3 + i − e^z) (e^z-1) = 0

that has a modulus smaller than 3π.

I'm thinking about using zero product property:

e^z - 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 $\displaystyle \begin{align*} z = \ln{2} + \frac{\pi}{6} i \end{align*}$
September 17th 2012, 08:17 AM
Tala

Re: Complex equation

This might be a stupid question but I can't see how the argument is π/6 ?
September 17th 2012, 08:20 AM
Tala

Re: Complex equation

I get it know:

arccos(√(3)/2) right ?
September 17th 2012, 08:23 AM
Prove It

Re: Complex equation

Quote:

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, $\displaystyle \begin{align*} \textrm{arg}\,(z) = \arctan{\left[ \frac{\textrm{Im}\,(z)}{\textrm{Re}\,(z)} \right]} \end{align*}$ .
September 17th 2012, 08:26 AM
Tala

Re: Complex equation

Thank you so much for the quick and brilliant answers.
I really appreciate it.
Tak (it means thank you in danish).
September 17th 2012, 08:41 AM
Plato

Re: Complex equation

Quote:

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.