1. ## Complex numbers question

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.

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

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

4. 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.

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

6. 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.
We use $\displaystyle \mbox{Im}(x+yi)=y$.
Then example: $\displaystyle \mbox{Im} (e^{i\pi } ) = \sin (\pi ) = 0$

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

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

9. No worries managed to work it out, thanks anyway.