$\blacktriangleright$ How do we calculate $(3 + 3\sqrt {3} \, i)^{\tfrac {5}{2}}$ ?

Trigonometric Thinking

Step 1: Convert the base to trigonometric form.

$$3 + 3\sqrt {3} \, i$$

$$= 6 \left ( \frac {1}{2} + \frac {\sqrt {3}}{2} \, i \right )$$

$$= 6 \left (\cos \left ( \frac {\pi }{3} \right )+i \, \sin \left ( \frac {\pi }{3} \right ) \right )$$

Step 2: The Angle

Find $\frac {\pi }{3}$ on the unit circle

We need to move the angle around.

First, the $\tfrac {1}{2}$ part of the exponent means we need $\tfrac {1}{2}$ of the angle: $\frac {\pi }{6}$.

Second, the $5$ in the exponent means we need to multiply the angle by $5$: $\frac {5\pi }{6}$.

Step 3: The Modulus

The $6$ also has the exponent. Here we think just in terms of real numbers: $6^{\tfrac {5}{2}}$.

Step 4: Reassemble

$$6^{\tfrac {5}{2}} \left (\cos \left ( \frac {5\pi }{6} \right )+i \, \sin \left ( \frac {5\pi }{6} \right ) \right )$$

$$= 6^{\tfrac {5}{2}} \left ( -\frac {\sqrt {3}}{2} + \frac {1}{2} \, i \right )$$

Exponential Thinking

Step 1: Convert the base to exponential form.

$$3 + 3\sqrt {3} \, i$$

$\sqrt {3^2 + (3\sqrt {3})^2} = \sqrt {9 + 9 \cdot 3} = \sqrt {36} = 6$

$\arctan \left ( \frac {3\sqrt {3}}{3} \right ) = \arctan \left ( \sqrt {3} \right ) = \frac {\pi }{3}$

$$= 6 \cdot e^{\tfrac {\pi }{3} \, i} = e^{\ln (6) + \tfrac {\pi }{3} \, i}$$

Step 2: The Exponent

$$= \left ( e^{\ln (6) + \tfrac {\pi }{3} \, i} \right )^{\tfrac {5}{2}}$$

$$= e^{\tfrac {5}{2} (\ln (6) + \tfrac {\pi }{3} \, i)}$$

$$= e^{\tfrac {5}{2} \ln (6) + \tfrac {5\pi }{6} \, i}$$

$$= 6^{\tfrac {5}{2}} \cdot e^{\tfrac {5\pi }{6} \, i}$$