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
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...
 
Mathematics 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.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top