Complex Numbers identity help

Click For Summary

Homework Help Overview

The discussion revolves around the manipulation of complex numbers, specifically focusing on the expression of the quotient of two complex numbers raised to a power. The original poster presents a problem involving the complex numbers z1 and z2, defined using trigonometric forms.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of trigonometric identities and the exponential form of complex numbers to simplify calculations. There are attempts to express the quotient of z1 and z2 and raise it to the third power. Some participants express uncertainty about their approaches and results.

Discussion Status

There is an ongoing exploration of different methods to tackle the problem. Some participants have provided guidance on using exponential forms, while others are sharing their attempts and expressing their confidence levels regarding the topic.

Contextual Notes

One participant mentions a looming unit test, indicating a time constraint and a need for clarity in understanding the topic. There is also an acknowledgment of uncertainty among participants regarding their solutions.

lunds002
Messages
21
Reaction score
0

Homework Statement



Let z1 = a (cos (pi/4) + i sin (pi/4) ) and z2 = b (cos (pi/3) + i sin (pi/3))
Express (z1/z2)^3 in the form z = x + yi.

]2. Homework Equations [/b]



The Attempt at a Solution



a(cos (pi/4) + i sin (pi/4))
b (cos (pi/3) + i sin (pi/3))

I then multiplied the top and bottom by the bottom reciprocal : cos (pi/3) - i sin (pi/3) to get

a(cos (pi/4) x cos (pi/3) + sin (pi/4) sin (pi/3)) + i (sin (pi/4) x cos (pi/3) - sin (pi/3) x cos (pi/4)
 
Physics news on Phys.org
Do you know the identity e=cosθ+isinθ?

It will make your calculations easier since you can express it in terms of the exponential function.
 
right, i forget that. thank you
 
I'm not sure if I'm right, I am stuck on this question as well.

I got z=(√2)a3/2b3 - (√2)a3/2b3 i
 
That's right.
 
vela said:
That's right.

That's great. I'm still kind of new to this topic, still not confident with it and I have an unit test coming up in less than a week's time! :eek:
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K