Question

I have z = 2cis(theta), what is -z in polar form?

Answer 1

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Answer 2

@AllTehMaffs ... could you help me?

Answer 3

yeah. Hi! So
\[ z =2 cis \theta = 2 \cos \theta + 2i\sin \theta\] right?

Answer 4

yea

Answer 5

factor out the 2 and it looks like
\[z = 2(\cos \theta + i \sin \theta) \]
Euler's formula is
\[\cos \theta + i \sin \theta = e^{i\theta}\]
and polar form looks like
\[z = re^{i \theta} \]
Soo.....

Answer 6

now that's something i have not learn yet. @@ i only started the basics of complex numbers.

Answer 7

You've not seen Euler's formula? hmmm... 1 sec...

What have you learned about so far?

Answer 8

yea.... i have not seen that formula.
I have learnt the basics of it, only \[z=rcis \theta\] multiplying and dividing in polar form, conjugate. That's all

Answer 9

okay, cool.. I'm just confused because becase 2cistheta *is* polar form.... so I don't know what they want you to convert it to >.<

could it just be
\[z = 2cis \theta
\\ \
\\ -z = -2cis \theta \]?
Do they want you to find theta?

Answer 10

hmmm.... they said that it can't be \[-z = -2cis \theta\] cuz if its written in \[z= rcis \theta\] , r needs to be positive. The answer is \[z=2cis(\theta+\pi)\] but no idea how they got that

Answer 11

OH! They are correct, and I totally forgot about that.
\[ -z = -2cis \theta = -2 \cos \theta -2 \sin \theta\]


\[ \cos (\theta + \pi) = - \cos \theta \]
\[ \sin( \theta + \pi) = - \sin \theta \]
Those are just trig identities. You can see they're true if you look at the (very poorly drawn) picture.. but using those

\[
\\ -z = 2 (-\cos \theta) +2 (-\sin \theta)
\\ \ \ \ = -2 \cos (\theta + \pi) -2 \sin (\theta + \pi)\
\\ \ \ \ = 2 \cos (\theta + \pi) +2 \sin (\theta + \pi)
\\
= 2 cis (\theta + \pi) \]
does that make sense?