absolute value of ((1+2i)/(1-2i))^2 ?? please..?

[(1 + 2i) / (1 - 2i)]²
= [(1 + 2i)² / (1 - 2i)(1 + 2i)]²
= [(1 + 4i + 4i²) / (1 - 4i²)]²
= [(1 + 4i - 4) / (1 + 4)]²
= [(-3 + 4i) / 5]²
= (-3 + 4i)² / 5²
= (9 - 24i + 16i²) / 25
= (9 - 24i - 16) / 25
= -(7/25) - (24/25)i

Absolute value of [(1 + 2i) / (1 - 2i)]²
= Absolute value of -(7/25) - (24/25)i
= √[(-7/25)² + (-24/25)²]
= 1

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

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

(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.