Question

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

49.${(1-i)}^5$

Short answer

${(1-i)}^5=-4+4i$.

Step-by-step solution

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

The magnitude of $1-i$ is

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

which gives

$$\begin{gathered} (1-i)=\sqrt{2}(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i) \\ =\sqrt{2}(\cos \frac{\mathrm{\pi }}{4}-i\sin \frac{\mathrm{\pi }}{4}) \end{gathered}$$

Step 2. Using De Moivre's Theorem.

We know that 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="1646763868779" $$\begin{gathered} {(1-i)}^5={\left[\sqrt{2}\right(\cos \frac{\mathrm{\pi }}{4}-i\sin \frac{\mathrm{\pi }}{4}\left)\right]}^5 \\ =\left[\right(\sqrt{2}{)}^5\left(\cos \right\{5\times \frac{\mathrm{\pi }}{4}\}-i\sin \{5\times \frac{\mathrm{\pi }}{4}\left\}\right)] \\ =4\sqrt{2}(\cos \frac{5\mathrm{\pi }}{4}-i\sin \frac{5\mathrm{\pi }}{4}) \\ =4\sqrt{2}(-\frac{1}{\sqrt{2}}-i(-\frac{1}{\sqrt{2}})) \\ =-4+4i \end{gathered}$$

So,${(1-i)}^5=-4+4i$