Polar Representation of a Complex Number

[Yankel]

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...

 

[MarkFL]

It might be simpler to use Euler's formula here:

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

And then, since:

\(\displaystyle e^{i\pi}=-1\)

We may conclude:

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

Likewise, since:

\(\displaystyle \overline{z}=re^{-i\theta}\)

Then:

\(\displaystyle -\overline{z}=re^{i(\pi-\theta)}\)

 

[Yankel]

Thank you, it's a very nice solution ! Is there another way of doing it, without Euler ?

 

[MarkFL]

Thank you, it's a very nice solution ! Is there another way of doing it, without Euler ?

Yes, if we consider the identities:

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

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

Then it follows that:

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

And if we consider the identities:

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

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

Then it follows that:

\(\displaystyle r\text{cis}(\pi-\theta)=-\overline{r\text{cis}(\theta)}\)

 

[Yankel]

Thank you !

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

Why is

\[rcis(360-\theta )=\bar{z}\] ?

 

[HallsofIvy]

The angle (argument) \(\displaystyle \theta\) 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.