Complex numbers question

**Johnt447** · October 31st 2008, 02:37 PM

Hey guys first post so be gentle

Came across this question here and really have no idea can someone help me out? Also am not sure how advanced it has to be to be in this section so I apologize if that's wrong. I just saw university level and since I'm at uni doing this I guessed would be a good place to put it.
Here's the question.

Thanks in advance.

**Plato** · October 31st 2008, 02:49 PM

$z = \sqrt 3 - i \Rightarrow \quad \arg (z) = - \frac{\pi }<br /> {6}\,,\,\left| z \right| = 2\,\& \,\mbox{Im} \left( {z^7 } \right) = \left| z \right|^7 \sin \left( {7\arg (z)} \right)$

**Johnt447** · October 31st 2008, 03:01 PM

Thanks for the reply

Any chance you can explain that to me I'm a bit confused.

**Plato** · October 31st 2008, 03:07 PM

Originally Posted by Johnt447

Any chance you can explain?

Well, I am not going to work this out for you.
You are the one who needs to learn to do these.

Just make the substitutions that I have suggested.
If that is still a mystery to you, I think you need to talk with your instructor.

**Johnt447** · October 31st 2008, 04:03 PM

I think i get it now. I'm not used to seeing Im, we use i in all the notes etc.

**Plato** · October 31st 2008, 06:29 PM

Originally Posted by Johnt447

I think i get it now. I'm not used to seeing Im, we use i in all the notes etc.

Your question is about finding the “imaginary part” of an expression!
We use $\mbox{Im}(x+yi)=y$ .
Then example: $\mbox{Im} (e^{i\pi } ) = \sin (\pi ) = 0$

**Johnt447** · November 1st 2008, 04:12 AM

Yeah i think that's me sorted, just not used to seeing it written like that.

**damid** · November 3rd 2008, 11:48 AM

hey john. actually that question is not that difficult. here goes. given that z=sqrt3-i. polar form is r[cos(theta)+isin(theta)] where r is the modulus of z. modulus of [z]=r=sqrta^6+b^2=2.(a=sqrt3 and b=-1) and the argument is arg(z)=-sqrt3/1=-60 so the polar form is after substituting what the unknows are and fix nicely; 2[cos(-60)+isin(-60)] NOTE that r is also the modulus or the distance. let me know wha you think

b).promise i'll get you b shortly

**Johnt447** · November 3rd 2008, 12:06 PM

No worries managed to work it out, thanks anyway.

**damid** · November 3rd 2008, 12:13 PM

i think this would be the answer for part b'. first work z^7 and that is equal; z^7=re^i(theta)=7[cos(theta)+isin(theta)]. but it is divided by 8 and the question wants the imaginary and the the real seperate. so
we 7[cos(theta)+isin(theta)]/8. now the real=7[cos(theta)/8 and Im=isin(theta)]/8 and solve your answer.looking forward to hear from you or anyone.

**Johnt447** · November 3rd 2008, 12:35 PM

Thanks for the reply that's what I done so I can rest easy

Thread: Complex numbers question

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