# complex number simplification

• Aug 8th 2012, 09:21 PM
linalg123
complex number simplification
Hi, sorry if this is the wrong forum, i did a search to see where people posted complex number stuff but it seemed to be a bit all over the place.

Write in polar form re^iθ

$\displaystyle \frac{2i}{3e^{4+i}}$

My problem is that in the denominator we have e^(4+i) and not e^iθ like usual. would it help splitting it into e^4 * e^i, i can't find a way to simplify it. (Headbang)(Headbang)(Headbang)
• Aug 8th 2012, 10:05 PM
Prove It
Re: complex number simplification
Quote:

Originally Posted by linalg123
Hi, sorry if this is the wrong forum, i did a search to see where people posted complex number stuff but it seemed to be a bit all over the place.

Write in polar form re^iθ

$\displaystyle \frac{2i}{3e^{4+i}}$

My problem is that in the denominator we have e^(4+i) and not e^iθ like usual. would it help splitting it into e^4 * e^i, i can't find a way to simplify it. (Headbang)(Headbang)(Headbang)

Do you know Euler's Formula? \displaystyle \displaystyle \begin{align*} e^{i\theta} \equiv \cos{\theta} + i\sin{\theta} \end{align*}. So

\displaystyle \displaystyle \begin{align*} \frac{2i}{3e^{4+i}} &= \frac{2\left(\cos{\frac{\pi}{2}} + i\sin{\frac{\pi}{2}}\right)}{3e^{4+i}} \\ &= \frac{2e^{\frac{\pi}{2}i}}{3e^{4+i}} \\ &= \frac{2}{3}e^{\frac{\pi}{2}i - \left(4 + i\right)} \\ &= \frac{2}{3}e^{-4 + \left(\frac{\pi}{2} - 1\right)i} \\ &= \frac{2}{3}e^{-4}e^{\left(\frac{\pi}{2} - 1\right)i} \end{align*}

So here we can see \displaystyle \displaystyle \begin{align*} r = \frac{2}{3}e^{-4} \end{align*} and \displaystyle \displaystyle \begin{align*} \theta = \frac{\pi}{2} - 1 \end{align*} :)