Find all solutions, z, of the equation,

$\displaystyle z^4 - z^3 + 27iz - 27i = 0$

expresing your answers in cartesian form.

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use the complex exponential to evaluate $\displaystyle \int e^{2t} sin(3t) dt$

Limits : 0 to pie/3

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- Jun 14th 2008, 05:18 PMpearlycFinding solutions for Complex Numbers
Find all solutions, z, of the equation,

$\displaystyle z^4 - z^3 + 27iz - 27i = 0$

expresing your answers in cartesian form.

---------------

use the complex exponential to evaluate $\displaystyle \int e^{2t} sin(3t) dt$

Limits : 0 to pie/3 - Jun 14th 2008, 05:19 PMtopsquark
- Jun 14th 2008, 05:22 PMtopsquark
I would have thought someone at your level would be able to spell "pi" correctly.

$\displaystyle \int_0^{\pi /3} e^{2t}sin(3t)~dt = Im \left ( \int_0^{\pi / 3}e^{2t}e^{3it}~dt \right )$

where Im( ) stands for the "imaginary part of."

Can you finish it from here?

-Dan - Jun 14th 2008, 06:18 PMpearlyc
okay, i got till ..

$\displaystyle Im \left ( \frac {(2-3i)(e^{(2+3i)t}}{13}\right )$

where do i go from here with the limits? - Jun 14th 2008, 06:28 PMmr fantastic
- Jun 14th 2008, 06:42 PMpearlyc
I converted exponential part back to the question and I got a zero :(

- Jun 14th 2008, 06:49 PMmr fantastic
- Jun 14th 2008, 06:57 PMpearlyc
$\displaystyle e^{(2+3i)t}$

coverted back to

$\displaystyle e^{2t} sin(3t)$

subbing $\displaystyle \frac{\pi}{3}$ makes $\displaystyle sin \pi = 0$ subbing it with 0 makes it = 0 too! - Jun 14th 2008, 08:16 PMmr fantastic
Is this the sort of work you'd present to your teacher? It's impossible from this 'working' to find your mistakes. Try again -

*all*the working this time please.

$\displaystyle \text{Im} \, \left ( \frac{2-3i}{13} \left[ (e^{(2+3i)t} \right]_{t = 0}^{t = \pi/3} \right ) = ....$