# Finding solutions for Complex Numbers

• Jun 14th 2008, 05:18 PM
pearlyc
Finding solutions for Complex Numbers
Find all solutions, z, of the equation,

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

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

use the complex exponential to evaluate $\int e^{2t} sin(3t) dt$
Limits : 0 to pie/3
• Jun 14th 2008, 05:19 PM
topsquark
Quote:

Originally Posted by pearlyc
Find all solutions, z, of the equation,

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

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

$z^3(z - 1) - 27i(z - 1) = 0$

$(z^3 - 27i)(z - 1) = 0$

$(z^3 + (3i)^3)(z - 1) = 0$
should give you a pretty good start.

-Dan
• Jun 14th 2008, 05:22 PM
topsquark
Quote:

Originally Posted by pearlyc
use the complex exponential to evaluate $\int e^{2t} sin(3t) dt$
Limits : 0 to pie/3

I would have thought someone at your level would be able to spell "pi" correctly.

$\int_0^{\pi /3} e^{2t}sin(3t)~dt = Im \left ( \int_0^{\pi / 3}e^{2t}e^{3it}~dt \right )$
where Im( ) stands for the "imaginary part of."

Can you finish it from here?

-Dan
• Jun 14th 2008, 06:18 PM
pearlyc
okay, i got till ..
$Im \left ( \frac {(2-3i)(e^{(2+3i)t}}{13}\right )$

where do i go from here with the limits?
• Jun 14th 2008, 06:28 PM
mr fantastic
Quote:

Originally Posted by pearlyc
okay, i got till ..
$Im \left ( \frac {(2-3i)(e^{(2+3i)t}}{13}\right )$

where do i go from here with the limits?

$\text{Im} \, \left ( \frac{2-3i}{13} \left[ (e^{(2+3i)t} \right]_{t = 0}^{t = \pi/3} \right ) = ....$
• Jun 14th 2008, 06:42 PM
pearlyc
I converted exponential part back to the question and I got a zero :(
• Jun 14th 2008, 06:49 PM
mr fantastic
Quote:

Originally Posted by pearlyc
I converted exponential part back to the question and I got a zero :(

The answer is not zero. You'll need to show all your working in order for your mistake(s) to be seen.
• Jun 14th 2008, 06:57 PM
pearlyc
$e^{(2+3i)t}$

coverted back to

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

subbing $\frac{\pi}{3}$ makes $sin \pi = 0$ subbing it with 0 makes it = 0 too!
• Jun 14th 2008, 08:16 PM
mr fantastic
Quote:

Originally Posted by pearlyc
$e^{(2+3i)t}$

coverted back to

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

subbing $\frac{\pi}{3}$ makes $sin \pi = 0$ subbing it with 0 makes it = 0 too!

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.

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