Let z = (1 + 2i)2 (1 – i)
= (1 + 4i + 4i2) (1 – i)
= [1 + 4i + 4(-1)] (1 – i) ….[∵ i = -1]
= (-3 + 4i) (1 – i)
= -3 + 3i + 4i – 4i2
= -3 + 7i – 4(-1)
= -3 + 7i + 4
∴ z = 1 + 7i
∴ a = 1, b = 7, i. e. a > 0, b > 0
∴ |z| = \(\sqrt{a^2+b^2}=\sqrt{1^2+7^2}=\sqrt{1+49}\)
= 5√2
Here, (1, 7) lies in 1st quadrant.
∴ amp(z) = tan-1\((\frac{b}a)\) = tan-17