Question

 (1 + 2i)² (1 – i)

Answer 1

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