Complex Numbers practice questions — Honors Math

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

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

$(1+i{)}^8=\text{\hspace{0.17em}⁣}{\left[\sqrt{2}\right]}^8\cdot \text{\hspace{0.17em}⁣}(\cos 2\pi +i\sin 2\pi )=16\cdot (1+0)=16$.

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

$\mathrm{\mid }z\mathrm{\mid }=\sqrt{1+3}=2$. Argument: $\tan \theta ={\displaystyle \frac{\sqrt{3}}{-1}}=-\sqrt{3}$, with $z$ in Q2, so $\theta ={\displaystyle \frac{2\pi }{3}}$. Hence $z=2\text{\hspace{0.17em}⁣}(\cos {\displaystyle \frac{2\pi }{3}}+i\sin {\displaystyle \frac{2\pi }{3}})$.

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

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

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

$z^2=-4\Rightarrow z^2+4=0\Rightarrow (z-2i)(z+2i)=0$. 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 ${\sum }_{k=0}^{n-1}{e}^{2\pi ik\mathrm{/}n}={\displaystyle \frac{1-1}{1-e^{2\pi i\mathrm{/}n}}}=0$ (for $n\ge 2$).