Author Topic: Complex number- Polar form (Read 4487 times) Tweet Share

Damo17 · « **on:** February 01, 2009, 09:43:44 pm »

Express $-2i cis \left (\theta -\frac{3\pi}{2} \right)$ in polar form.

Any help would be appreciated.

Mao · « **Reply #1 on:** February 01, 2009, 10:32:39 pm »

/0 · « **Reply #2 on:** February 02, 2009, 06:42:06 am »

$-2i \mbox{cis}=2\mbox{cis}\left(-\frac{\pi}{2}\right)$

Can this only be done if there is no cosine term in the cis?

Mao · « **Reply #3 on:** February 02, 2009, 07:56:59 pm »

wha? That didn't make sense.

$-2i = 2\mbox{cis} \left( -\frac{\pi}{2}\right)$
these two are equivalent, on the left is the cartesian expression, on the right is the polar form.

Not sure why you included cis on the LHS, by itself (without an argument), cis does not have meaning, much like how 'sin' has no meaning unless it's sin of something.

/0 · « **Reply #4 on:** February 02, 2009, 08:39:09 pm »

Sorry I meant $-2i = 2\mbox{cis}\left(-\frac{\pi}{2}\right)$

Mao · « **Reply #5 on:** February 02, 2009, 08:46:28 pm »

I still do not understand the nature of the question..

on the Argand diagram, -i points vertically downwards, which is the argument $-\frac{\pi}{2}$ .

/0 · « **Reply #6 on:** February 02, 2009, 09:07:31 pm »

I mean for instance say

$-2\mbox{cis}\left(\theta - \frac{\pi}{3}\right) = 2\mbox{cis}\left(\frac{\pi}{3} - \theta}\right)$

Is that not true in general?

Mao · « **Reply #7 on:** February 03, 2009, 12:25:07 am »

no. that rule is true for sine and tangent, but not cis.
e.g.
$-\sin (\theta) = \sin (-\theta)$

For the case of $\mbox{cis}(\theta)$ , or more conventionally written as $e^{i\theta}$ as per Euler's formula, the effect of negative in front is to add pi to the argument:

$-\mbox{cis}(\theta) = -1 \times \mbox{cis}(\theta) = \mbox{cis}(\pi) \times \mbox{cis}(\theta) = \mbox{cis}(\theta + \pi)$

Damo17 · « **Reply #8 on:** February 19, 2009, 08:30:58 pm »

Find the solutions of the equation $z^4-2z^2+4=0$ in polar form.

danieltennis · « **Reply #9 on:** February 19, 2009, 08:39:37 pm »

Hint: $Let$ $z^2 = a$

Damo17 · « **Reply #10 on:** February 19, 2009, 08:41:32 pm »

Quote from: daniel. on February 19, 2009, 08:39:37 pm

Hint: $Let$ $z^2 = a$

Oh, totally forgot about that way. Thanks

Damo17 · « **Reply #11 on:** February 20, 2009, 01:20:19 pm »

If the real part of $\frac{z+1}{z-1}$ is zero, find the locus of points representing z in the complex plane.
Any ideas on what to do first?

shinny · « **Reply #12 on:** February 20, 2009, 03:43:01 pm »

This question is pretty ugly so hopefully I didn't make any mistakes.

Let $z=x+iy$

$\frac{z+1}{z-1}=\frac{x+1+iy}{x-1+iy}$

$=\frac{x+1+iy}{x-1+iy} \times \frac{x-1-iy}{x-1-iy}$

$=\frac{[(x+1)+iy][(x-1)-iy]}{(x-1)^{2}-(iy)^{2}}$

$=\frac{(x+1)(x-1)+y^{2}}{(x-1)^{2}+y^{2}}+\frac{(x-1)iy-(x+1)iy}{(x-1)^{2}+y^{2}}$

$\because$ $Re(\frac{z+1}{z-1})=\frac{(x+1)(x-1)+y^{2}}{(x-1)^{2}+y^{2}}=0$

$\therefore$ $\frac{z+1}{z-1}=\frac{(x-1)iy-(x+1)iy}{(x-1)^{2}+y^{2}}$

$=-\frac{2y}{(x-1)^{2}+y^{2}}$

DISCLAIMER: I'd say there's a 70% chance of an error in there in some form or another, particularly in my latex'ing =P Try something similar I guess and see if you end up with the same answer.

Mao · « **Reply #13 on:** February 20, 2009, 08:24:35 pm »

yeah... $\frac{1}{z} = \frac{1}{z} \times \frac{\bar{z}}{\bar{z}} = \frac{\bar{z}}{|z|^2}$

sorry shinny, your expansion in the first step would give $(x-1)^2-y^2 - 2(x-1)y i$ as the denominator.

shinny · « **Reply #14 on:** February 20, 2009, 08:37:02 pm »

Knew it =P Always get errors when I go straight to latexing.

EDIT: Oh it was only a minor error; I just typed the wrong sign there but I did the expansion as intended so it didn't affect the final outcome. Any other mistakes? =\

EDIT 2: On another note, is a x-y relationship even what I was meant to find? ._. Or is there some sort of way to leave this simpler.

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