Complex Numbers: Numerical Problems Set (7)

$z = 1 + i$. Square first:

$z^2 = (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$

Square again:

$z^4 = (2i)^2 = 4i^2 = -4$

$|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$

$\arg z = \arctan\frac{1}{1} = \frac{\pi}{4}$

(Since $z$ is in the first quadrant, no adjustment needed.)

$z = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) = \sqrt{2}\, e^{i\pi/4}$

Verify by De Moivre’s: $z^4 = (\sqrt{2})^4 e^{i\pi} = 4(\cos\pi + i\sin\pi) = -4$. ✓ Matches step 1.

Write $-16$ in polar: $-16 = 16 e^{i\pi} = 16 e^{i(\pi + 2k\pi)}$ for integer $k$.

Fourth roots:

$w_k = 16^{1/4} e^{i(\pi + 2k\pi)/4} = 2 e^{i(2k+1)\pi/4}, \quad k = 0, 1, 2, 3$

Compute each:

• $k=0$: $2 e^{i\pi/4} = 2(\cos 45° + i\sin 45°) = \sqrt{2} + \sqrt{2}\,i$
• $k=1$: $2 e^{i 3\pi/4} = -\sqrt{2} + \sqrt{2}\,i$
• $k=2$: $2 e^{i 5\pi/4} = -\sqrt{2} - \sqrt{2}\,i$
• $k=3$: $2 e^{i 7\pi/4} = \sqrt{2} - \sqrt{2}\,i$

The four roots sit at the vertices of a square in the complex plane, each at distance $2$ from the origin.