Anonymous

Absolute valeu of ((1 + 2i)/(1 - 2i))^2 = abs((1 + 2 i)/(1 - 2 i))^2 = 1

Anonymous

Let 1 = r cos theta 2 = r sin theta r= sqrt( 1+4) = sqrt 5 theta = tan ⁻¹ ( 2/1) So we have 1+2i = r exp +i theta 1- 2 i = r exp - i theta (1+2i)/(1-2i) = exp 2i theta ((1+2i)/(1-2i) ) = exp (4i theta) Absolute value =1

Anonymous

(1+2i)/(1-2i) = (1+2i)^2/(1+4) = (4i - 3)/5. The magnitude of this (without squaring) is sqrt(5)/5, so after you square it you get 1/5. And that's the answer.