trig representation to find argument and modulus

Oct 2012
15
0
hi again guys!!

new problem now..

question:

z=((4+4jsqrt3)/(-6+6j))^2


so this time i was thinking to multiply by the conjugate and then i can use modulus= sort a^2 +b^2 and arctan(b/a)=argument... then use de moivres theorem to deal with the square... im just reading the text book and trying to get what I can as I started the course late :(

thanks for any help :)
 

ZardoZ

Math Team
Nov 2010
1,990
133
Greece, Thessaloniki
\(\displaystyle \left[\frac{4+4\sqrt{3}i}{-6+6i}\right]^{2}=\left[ \frac{4(1+\sqrt{3}i)}{6(-1+i)}\right]^{2}=\left(\frac{4}{6}\right)^{2}\cdot\frac{(1+\sqrt{3}i)^{2}}{(-1+i)^{2}}=\frac{4}{9}\cdot \frac{4 \left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)^{2}}{2\left(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)^{2}}=\frac{8}{9}\cdot\left(\frac{e^{\frac{\pi}{3}i}}{e^{\frac{3\pi}{4}i}}\right)^{2}=\fbox{\frac{8}{9} }e^{\fbox{-\frac{5\pi}{6}}i}=-\frac{8}{9}e^{\frac{\pi}{6}i}=-\frac{8}{9}\left(\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right)\)