Question

In Problems 41–52, write each expression in the standard form $a+bi$.

$50.{(\sqrt{3}-i)}^6$

Short answer

${(\sqrt{3}-i)}^6=-64$

Step-by-step solution

Step 1. Rewrite (3-i) in the form r(cos∅+isin∅).

The magnitude of $(\sqrt{3}-i)$is

$$\begin{gathered} \sqrt{{\left(\sqrt{3}\right)}^2+{(-1)}^2}=\sqrt{3+1} \\ =\sqrt{4} \\ =2 \end{gathered}$$

which gives

$$\begin{gathered} (\sqrt{3}-i)=2(\frac{\sqrt{3}}{2}-i\frac{1}{2}) \\ =2(\cos 330°-i\sin 330°) \end{gathered}$$

Step 2. Using De Moivre's Theorem.

If a complex number is $z=r(\cos \varnothing +i\sin \varnothing )$then $z^n=r^n\left[\cos \right(n\varnothing )+i\sin (n\varnothing \left)\right]$.

This gives

role="math" localid="1646805734516" $$\begin{gathered} {(\sqrt{3}-i)}^6={\left[2\right(\cos 330°-i\sin 330°\left)\right]}^6 \\ =\left\{\right(2{)}^6\left[\cos \right(6\times 330°)-i\sin (6\times 330°\left)\right]\} \\ =64(\cos 1980°-i\sin 1980°) \\ =64(\cos 180°-i\sin 180°) \\ =64(-1+0i) \\ =-64 \end{gathered}$$

So${(\sqrt{3}-i)}^6=-64$