MHB Polar Representation of a Complex Number

AI Thread Summary
The discussion focuses on finding the polar representation of the complex numbers -z and -z̅. The correct representations are identified as rcis(180 + θ) for -z and rcis(180 - θ) for -z̅, derived using Euler's formula and trigonometric identities. The participants confirm that the polar form of the conjugate z̅ is rcis(-θ), utilizing the properties of cosine and sine functions. Additionally, the relationship rcis(360 - θ) = z̅ is explained through the transformation of angles and the even-odd nature of the trigonometric functions. The conversation concludes with a clear understanding of these polar representations and their derivations.
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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