Question

The modulus and amplitude of $\frac{ 1 + 2i}{1 - (1 - i)^2}$ are respectively

• A. $1, \frac{\pi}{3}$
• B. $\sqrt{2} , \frac{\pi}{6}$
• C. $1, 0$
• D. $\sqrt{3} , 0$

Answer 1

Answer: (C)

$1, 0$

Answer 2

Let $Z = \frac{ 1 + 2i}{1 - (1 - i)^2}$
$= \frac{1+2i}{1-\left(1 + i^{2} - 2i\right)} = \frac{1+2i}{1+2i} = 1 \Rightarrow Z = 1 + 0i$
Now, put $1 = r \cos \theta , 0 = r \sin \theta$
$r= \sqrt{1+0} = 1 \Rightarrow \left|Z\right| = \sqrt{a^{2} + b^{2}} = \sqrt{1^{2}+0} = 1$
$\therefore \:\:\: \tan \theta = 0 \Rightarrow \theta = 0$
$\therefore$ Modulus is 1 and amplitude is 0.