How Do You Solve Complex Number Equations with Trigonometric Forms?

[VADER25]

hi, I am trying to solve this equation and i would like some help.
i've done some of it already and i don't know how to go on from here.

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


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

 

[berkeman]

It looks like you multiplied out (-i)(4-4i) incorrectly...?

 

[HallsofIvy]

hi, I am trying to solve this equation and i would like some help.
i've done some of it already and i don't know how to go on from here.

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


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 $z= r(cos(\theta)+ i sin(\theta))= r e^{i\theta}$.