Complex numbers

  • Thread starter VADER25
  • Start date
  • #1
6
0
hi, im trying to solve this equation and i would like some help.
i've done some of it already and i dont know how to go on from here.

[tex]z=-\frac{(4-4i)(\sqrt{6}-i\sqrt{2})}{i}\\ =-\frac{(4-4i)(\sqrt{6}-i\sqrt{2})-i}{i(-i)}=(4i+4)(\sqrt{6}-i\sqrt{2})\\ \hspace{6} r=\sqrt{4^{2}+4^{2}}=\sqrt{32}\hspace{6} and \hspace{6} v=\frac{\pi }{4}\hspace{6}[/tex]


the answer should be in this form a*pi/b
 
Last edited:

Answers and Replies

  • #2
berkeman
Mentor
61,252
11,755
It looks like you multiplied out (-i)(4-4i) incorrectly....?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
967
hi, im trying to solve this equation and i would like some help.
i've done some of it already and i dont know how to go on from here.

[tex]z=-\frac{(4-4i)(\sqrt{6}-i\sqrt{2})}{i}\\ =-\frac{(4-4i)(\sqrt{6}-i\sqrt{2})-i}{i(-i)}=(4i+4)(\sqrt{6}-i\sqrt{2})\\ \hspace{6} r=\sqrt{4^{2}+4^{2}}=\sqrt{32}\hspace{6} and \hspace{6} v=\frac{\pi }{4}\hspace{6}[/tex]


the answer should be in this form a*pi/b
What exactly is the question then? You gave an expression of z and now you want the "argument" of z? Any complex number can be written in the form [itex]z= r(cos(\theta)+ i sin(\theta))= r e^{i\theta}[/itex].
 

Related Threads on Complex numbers

  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
1
Views
788
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
992
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
952
  • Last Post
Replies
6
Views
586
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
1
Views
1K
Top