Question

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

$51.{(\sqrt{2}-i)}^6$

Short answer

${(\sqrt{2}-i)}^6=-23+14.14i$

Step-by-step solution

Step 1. Rewrite a+bi as r(cos∅+isin∅).

The magnitude of $(\sqrt{2}-i)$is

role="math" localid="1646806173217" $$\begin{gathered} \sqrt{{\left(\sqrt{2}\right)}^2+{(-1)}^2}=\sqrt{2+1} \\ =\sqrt{3} \end{gathered}$$

This gives

$$\begin{gathered} (\sqrt{2}-i)=\sqrt{3}(\frac{\sqrt{2}}{\sqrt{3}}-i\frac{1}{\sqrt{3}}) \\ =\sqrt{3}(\cos 35.264°+i\sin 35.264°) \end{gathered}$$

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="1646806473235" $$\begin{gathered} {(\sqrt{2}-i)}^6={\left[\sqrt{3}\right(\cos 35.264°)-i\sin (35.264°\left)\right]}^6 \\ =\left\{\right(\sqrt{3}{)}^6\left[\cos \right(6\times 35.264)-i\sin (6\times 35.264\left)\right]\} \\ =27(\cos 211.584°-i\sin 211.584°) \\ =27(-0.8518-i(-0.8518)) \\ =-23+14.14i \end{gathered}$$

So,${(\sqrt{2}-i)}^6=-23+14.14i$