MHB Polar Representation of a Complex Number

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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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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