Complex Numbers practice questions — Honors Math

Since $13 = 4\cdot 3 + 1$, we have $$i^{13} = i^{4\cdot 3 + 1} = (i^4)^3 \cdot i = i.$$ So $z = 4 + 3i$, and $$|z| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5.$$

$\dfrac{1 + i}{1 - i} \cdot \dfrac{1 + i}{1 + i} = \dfrac{(1 + i)^2}{1 - i^2} = \dfrac{2i}{2} = i$.

$(1 + i)^8 = \!\left[\sqrt{2}\right]^{8} \cdot \!\left(\cos 2\pi + i\sin 2\pi\right) = 16 \cdot (1 + 0) = 16$.

By the Fundamental Theorem of Algebra (or Moivre), $z^n = 1$ has exactly $n$ distinct complex roots, the $n$th roots of unity $\varepsilon_k = e^{2\pi i k/n}$, $k = 0, \ldots, n-1$.

$|z| = \sqrt{1 + 3} = 2$. Argument: $\tan\theta = \dfrac{\sqrt{3}}{-1} = -\sqrt{3}$, with $z$ in Q2, so $\theta = \dfrac{2\pi}{3}$. Hence $z = 2\!\left(\cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3}\right)$.

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

$\overline{a + bi} = a - bi$, so $\overline{2 - 3i} = 2 + 3i$.

$|z| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

$z^2 = -4 \Rightarrow z^2 + 4 = 0 \Rightarrow (z - 2i)(z + 2i) = 0$. Solutions: $z = 2i$ or $z = -2i$.

The polynomial $z^n - 1$ has zero coefficient for $z^{n-1}$, so by Viète's formula the sum of its roots is $0$. Equivalently, the geometric sum $\sum_{k=0}^{n-1} e^{2\pi i k / n} = \dfrac{1 - 1}{1 - e^{2\pi i / n}} = 0$ (for $n \ge 2$).