Answer to Question #219926 in Linear Algebra for Pocupine

Question

Find an expression of z^n which nEN

Answer 1

$z =\frac{z_1}{z_2} \\ z_1 = tan θ + i \\ z_2 = \bar{z_1} = \bar{(\tan\theta+i)}=tan \theta -i\\ z =\frac{z_1}{z_2} =\frac{tan \theta +i}{tan \theta -i}\\ z =\frac{tan \theta +i}{tan \theta -i}*\frac{tan \theta +i}{tan \theta +i}\\ Simplifying \\ z= \frac{tan^2 \theta +i^2+2 tan \theta i}{tan^2 \theta +1}\\ z= \frac{tan^2 \theta -1+2 tan \theta i}{sec^2 \theta}\\ z= \frac{\frac{sin^2 \theta}{cos^2 \theta} -1+2 \frac{sin \theta}{cos \theta}i}{\frac{1}{cos^2 \theta}}\\ z= \frac{sin^2 \theta-cos^2 \theta+2sin \theta cos \theta i}{\frac{1}{cos^2 \theta}*cos^2\theta}\\ z=- cos 2\theta + sin2 \theta i\\ Hence\\ e^{i \theta}= cos \theta + isin \theta\\ e^{-i \theta}= cos \theta - isin \theta\\ \implies e^{i2\theta}= cos 2\theta + isin 2\theta\\ \space \space \space \space \space \space \space \space \space \space e^{-i \theta}= cos 2\theta -isin2 \theta\\ Adding \space them\\ cos 2 \theta = \frac{e^{i2 \theta }-e^{-i2 \theta }}{2}\\ Also, sin 2 \theta = \frac{e^{i2 \theta }-e^{-i2 \theta }}{2i}\\ z= \frac{e^{i2 \theta }-e^{-i2 \theta }}{2}+ \frac{e^{i2 \theta }-e^{-i2 \theta }}{2i}\\ z= -e^{i2 \theta }\\ \therefore z^n= (-e^{i2 \theta })^n = (-)^ne^{i2n \theta }$