Exponential form and powers of complex numbers

Problem

Let $z=1+\sqrt3\,i$ and $w=-1+i$.

(a) Write $z$ and $w$ in exponential form (modulus and argument) and sketch the corresponding vectors.

(b) Compute $z^{4}$ in algebraic form.

(c) Compute $1/w^{6}$ in algebraic form.

Solution

(a) The modulus of $z$ is $|z|=\sqrt{1^2+(\sqrt3)^2}=2$, and since $(1,\sqrt3)$ lies in the first quadrant, $\arg z=\dfrac\pi3$, so $z=2e^{i\pi/3}$. For $w$, $|w|=\sqrt{(-1)^2+1^2}=\sqrt2$ and, since $(-1,1)$ lies in the second quadrant, $\arg w=\dfrac{3\pi}4$, so $w=\sqrt2\,e^{i3\pi/4}$. The two vectors have length $2$ and $\sqrt2$ and make angles of $60°$ and $135°$ with the positive real axis, respectively.

(b) Using the exponential form, $z^4=2^4e^{i4\pi/3}=16e^{i4\pi/3}=16\left(-\tfrac12-\tfrac{\sqrt3}2 i\right)=-8-8\sqrt3\,i.$

(c) We have $w^6=(\sqrt2)^6e^{i6\cdot3\pi/4}=8e^{i9\pi/2}$. Since $9\pi/2$ and $\pi/2$ differ by a multiple of $2\pi$ (indeed $9\pi/2-4\pi=\pi/2$), $w^6=8e^{i\pi/2}=8i$. Hence $\dfrac1{w^6}=\dfrac1{8i}=-\dfrac i8.$

$\boxed{z=2e^{i\pi/3},\quad w=\sqrt2\,e^{i3\pi/4},\quad z^4=-8-8\sqrt3\,i,\quad \frac1{w^6}=-\frac i8}$