Question B.5: If A = 6 + j8, find: (a) √A, (b) A^4.

(a) First, convert A to polar form:

r=\sqrt{6^{2}+8^{2}}=10, \quad \theta=\tan ^{-1} \frac{8}{6}=53.13^{\circ}, \quad A=10 \angle 53.13^{\circ}

Then

\sqrt{A}=\sqrt{10} \angle 53.13^{\circ} / 2=3.162 \angle 26.56^{\circ}

(b) Since A = 10∠53.13°,

A^{4}=r^{4} \angle 4 \theta=10^{4} \angle 4 \times 53.13^{\circ}=10,000 \angle 212.52^{\circ}