Problem Statement
Convert the complex number $z = \frac{1 + 7i}{(2 - i)^2}$ into standard polar form $r(\cos\theta + i\sin\theta)$.
Verified Solution & Marking Scheme
Simplify the denominator (2 - i)²
$$(2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$$
Multiply numerator and denominator by conjugate
$$z = \frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i} = \frac{(1 + 7i)(3 + 4i)}{3^2 + 4^2}$$
$$= \frac{3 + 4i + 21i + 28i^2}{9 + 16} = \frac{3 + 25i - 28}{25} = \frac{-25 + 25i}{25} = -1 + i$$
Determine modulus r and principal argument θ
$$r = |z| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$$
Since $x = -1 < 0$ and $y = 1 > 0$, $z$ lies in the second quadrant.
$$\alpha = \tan^{-1}\left|\frac{1}{-1}\right| = \frac{\pi}{4} \implies \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$
Polar form: $z = \sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)$.