Question

find

Answer 1

\[\left| \dfrac{2e^{i \theta} - i}{ie^{i \theta}+2} \right|\]

Answer 2

we may use the property \(\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}\) :

\[\left| \dfrac{2e^{i \theta} - i}{ie^{i \theta}+2} \right| = \dfrac{|2\cos\theta + i(2\sin\theta-1)|}{|2-\sin\theta+i\cos\theta|}=1\]

Answer 3

yep!