The Complex Exponential And Regions in the Complex Plane

• Sep 9th 2011, 10:27 PM
andrew2322
The Complex Exponential And Regions in the Complex Plane
Hello all, I have done a few questions that I need corrected.

1.) Sketch the region of the complex plane defined by Im (z^2) = 2 for pi/8 < arg (z) < 3pi/8

my solution: I let z = x + iy, then found z^2 and the imaginary parts of z^2 which gave me a graph shape of 1/x. the region defined by the angle is in the first quadrant and only a small bit of that curve is included.

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2.) z = 1/4 - [root(3) / 4]i

a) Write z in exponential polar form for angles between -pi and pi

my solution: 1/2 * e ^ i (pi/3)

b) Calculate (1/z) ^ 9

my solution: -2 * 2^8 + 0i

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3.) Use the complex exponential to express sin^3(theta) as a sum of sines or cosines of multiples of theta.

My solution: -1/i sin(3 theta) + 3/i cos (theta)
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• Sep 9th 2011, 10:46 PM
Prove It
Re: The Complex Exponential And Regions in the Complex Plane
Quote:

Originally Posted by andrew2322
Hello all, I have done a few questions that I need corrected.

1.) Sketch the region of the complex plane defined by Im (z^2) for pi/8 < arg (z) < 3pi/8

my solution: I let z = x + iy, then found z^2 and the imaginary parts of z^2 which gave me a graph shape of 1/x. the region defined by the angle is in the first quadrant and only a small bit of that curve is included.

__________________________________________________ ______________________

2.) z = 1/4 - [root(3) / 4]i

a) Write z in exponential polar form for angles between -pi and pi

my solution: 1/2 * e ^ i (pi/3)

b) Calculate (1/z) ^ 9

my solution: -2 * 2^8 + 0i

__________________________________________________ ______________________

3.) Use the complex exponential to express sin^3(theta) as a sum of sines or cosines of multiples of theta.

My solution: -1/i sin(3 theta) + 3/i cos (theta)
__________________________________________________ ______________________

1. Are you sure it didn't say $\displaystyle \displaystyle \frac{\pi}{8} < \arg{\left(z^2\right)} < \frac{3\pi}{8}$

2. a) It's clear that $\displaystyle \displaystyle z = \frac{1}{4} - \frac{\sqrt{3}}{4}i$ is in the fourth quadrant, so $\displaystyle \displaystyle \arg{(z)}$ can not possibly be $\displaystyle \displaystyle \frac{\pi}{3}$.

b) This is easiest to evaluate if you have evaluated part a) correctly...
• Sep 10th 2011, 12:18 AM
andrew2322
Re: The Complex Exponential And Regions in the Complex Plane
Hello Prove it. I checked the question again and it says arg (z) not arg (z^2)

also, for part 2a) I realise my careless mistake, I was meant to type - pi/3.

due apologies.
• Sep 10th 2011, 02:49 AM
andrew2322
Re: The Complex Exponential And Regions in the Complex Plane
Sorry Prove It and others. The Im (z^2) = 2, i forgot to include the = 2 part. very sorry.

I have edited the original post to include this correction.
• Sep 10th 2011, 06:56 AM
Prove It
Re: The Complex Exponential And Regions in the Complex Plane
Quote:

Originally Posted by andrew2322
Hello Prove it. I checked the question again and it says arg (z) not arg (z^2)

also, for part 2a) I realise my careless mistake, I was meant to type - pi/3.

due apologies.

Then $\displaystyle \displaystyle z = \frac{1}{2}e^{-\frac{\pi i}{3}}$ is correct.
• Sep 10th 2011, 04:51 PM
andrew2322
Re: The Complex Exponential And Regions in the Complex Plane
is part B correct?
• Sep 10th 2011, 08:03 PM
Prove It
Re: The Complex Exponential And Regions in the Complex Plane
Quote:

Originally Posted by andrew2322
is part B correct?

You tell me. $\displaystyle \displaystyle z = \frac{1}{2}e^{-\frac{\pi i}{3}}$, so what is $\displaystyle \displaystyle z^{-9}$?
• Sep 10th 2011, 11:49 PM
andrew2322
Re: The Complex Exponential And Regions in the Complex Plane
2^9 e ^3pi?
• Sep 10th 2011, 11:51 PM
andrew2322
Re: The Complex Exponential And Regions in the Complex Plane
which is equivalent to 2^9 e^pi which is 2^9 (cos(pi) +isin(pi)) which is -1 * 2^9?
• Sep 11th 2011, 12:05 AM
Siron
Re: The Complex Exponential And Regions in the Complex Plane
Quote:

Originally Posted by andrew2322
2^9 e ^3pi?

Correct, but don't forget the 'i':
$\displaystyle \left(\frac{1}{2}e^{-\frac{\pi i}{3}}\right)^{-9}=2^9e^{3\pi i}$
• Sep 11th 2011, 12:06 AM
andrew2322
Re: The Complex Exponential And Regions in the Complex Plane
Cool, thanks for the heads up. I really have to stop doing that haha.

Would you happen to have any clue on how to do Q1)? I reposted it because this one got quite messy and still have yet to receive a response.