Polar Representation of a Complex Number

Click For Summary

Discussion Overview

The discussion revolves around the polar representation of complex numbers, specifically focusing on the representations of \(-z\) and \(-z\bar{}\) given a complex number in polar form. Participants explore various methods to derive these representations, including trigonometric identities and Euler's formula.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents the polar form of a complex number \(z\) as \(z=r(\cos\theta + i\sin\theta)\) and seeks to find the polar representations of \(-z\) and \(-z\bar{}\).
  • Another participant suggests using Euler's formula, stating that \(-z\) can be expressed as \(re^{i(\pi+\theta)}\) and \(-\overline{z}\) as \(re^{i(\pi-\theta)}\).
  • A participant questions whether there is an alternative method to derive the results without using Euler's formula.
  • Further identities involving trigonometric functions are discussed, where participants note that \(\cos(\pi+\theta)=-\cos(\theta)\) and \(\sin(\pi+\theta)=-\sin(\theta)\) to derive the polar forms.
  • Another participant inquires about the relationship between \(rcis(360-\theta)\) and \(\bar{z}\), leading to a discussion on how the angle transformation relates to the complex conjugate.
  • One participant explains that \(360-\theta\) corresponds to the negative angle, leading to the conclusion that \(rcis(360-\theta) = \bar{z}\) based on the properties of sine and cosine functions.

Areas of Agreement / Disagreement

Participants express various methods and identities to derive the polar representations, but there is no consensus on a single approach. Multiple competing views and methods remain present throughout the discussion.

Contextual Notes

Some participants rely on specific trigonometric identities and properties of functions, which may depend on the definitions and assumptions made about angles and complex numbers. The discussion does not resolve the mathematical steps fully, leaving some aspects open to interpretation.

Yankel
Messages
390
Reaction score
0
Hello all,

Given a complex number:

\[z=r(cos\theta +isin\theta )\]

I wish to find the polar representation of:

\[-z,-z\bar{}\]

I know that the answer should be:

\[rcis(180+\theta )\]

and

\[rcis(180-\theta )\]

but I don't know how to get there. I suspect a trigonometric identity, but I couldn't figure it out.

I did manage to fine that the polar representation of

\[z\bar{}\]

is

\[rcis(-\theta )\]

but I did that using the fact that cos is an even function and sin is odd.

Thank you !

- - - Updated - - -

z- is the conjugate, I don't know why my Latex went so wrong...
 
Physics news on Phys.org
It might be simpler to use Euler's formula here:

$$z=r\text{cis}(\theta)=re^{i\theta}$$

And then, since:

$$e^{i\pi}=-1$$

We may conclude:

$$-z=re^{i\pi}e^{i\theta}=re^{i(\pi+\theta)}$$

Likewise, since:

$$\overline{z}=re^{-i\theta}$$

Then:

$$-\overline{z}=re^{i(\pi-\theta)}$$
 
Thank you, it's a very nice solution ! Is there another way of doing it, without Euler ?
 
Yankel said:
Thank you, it's a very nice solution ! Is there another way of doing it, without Euler ?

Yes, if we consider the identities:

$$\cos(\pi+\theta)=-\cos(\theta)$$

$$\sin(\pi+\theta)=-\sin(\theta)$$

Then it follows that:

$$r\text{cis}(\pi+\theta)=-r\text{cis}(\theta)$$

And if we consider the identities:

$$\cos(\pi-\theta)=-\cos(\theta)$$

$$\sin(\pi-\theta)=\sin(\theta)$$

Then it follows that:

$$r\text{cis}(\pi-\theta)=-\overline{r\text{cis}(\theta)}$$
 
Thank you !

May I ask something related (therefore won't open a new thread for it).

Why is

\[rcis(360-\theta )=\bar{z}\] ?
 
The angle (argument) $$\theta$$ is measured from the real (x) axis. 360- \theta (I would say 2\pi- \theta) changes from \theta to -\theta so r(cos(\theta)+ i sin(\theta)) to r(cos(-\theta)+ i sin(-\theta)), which, because cosine is an "even function" (cos(-\theta)= cos(\theta)) and sine is an "odd function" (sin(-\theta)= -sin(\theta)), equals r(cos(\theta)- i sin(\theta)), the complex conjugate.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K