MHB Polar Representation of a Complex Number

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SUMMARY

The discussion focuses on the polar representation of complex numbers, specifically how to derive the polar forms of \(-z\) and \(-\overline{z}\) from the original complex number \(z = r(\cos \theta + i \sin \theta)\). The correct polar representations are established as \(rcis(180 + \theta)\) for \(-z\) and \(rcis(180 - \theta)\) for \(-\overline{z}\). The participants utilize trigonometric identities and Euler's formula to derive these results, confirming the relationships between angles and their corresponding sine and cosine values.

PREREQUISITES
  • Understanding of complex numbers and their polar forms
  • Familiarity with Euler's formula: \(e^{i\theta} = \cos \theta + i \sin \theta\)
  • Knowledge of trigonometric identities, particularly for sine and cosine functions
  • Basic algebraic manipulation of complex expressions
NEXT STEPS
  • Study the derivation of complex conjugates in polar form
  • Explore advanced trigonometric identities and their applications in complex analysis
  • Learn about the geometric interpretation of complex numbers on the Argand plane
  • Investigate the applications of Euler's formula in electrical engineering and signal processing
USEFUL FOR

Mathematicians, physics students, engineers, and anyone interested in understanding complex number representations and their applications in various fields.

Yankel
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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...
 
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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.
 

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