Simplify 1/sqrt2(cos45^@-isin45^@)^5 in the form a+ib using De Moivre's theorem?

1 Answer
Nov 15, 2017

$\frac{1}{\sqrt{2}} {\left(\cos {45}^{\circ} - i \sin {45}^{\circ}\right)}^{5} = - \frac{1}{2} + i \frac{1}{2}$

Explanation:

According to De Moivre's theorem ${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

Hence $\frac{1}{\sqrt{2}} {\left(\cos {45}^{\circ} - i \sin {45}^{\circ}\right)}^{5}$

= $\frac{1}{\sqrt{2}} {\left(\cos \left(- {45}^{\circ}\right) + i \sin \left(- {45}^{\circ}\right)\right)}^{5}$

= $\frac{1}{\sqrt{2}} \left(\cos \left(- 5 \times {45}^{\circ}\right) + i \sin \left(- 5 \times {45}^{\circ}\right)\right)$

= $\frac{1}{\sqrt{2}} \left(\cos \left(- {225}^{\circ}\right) + i \sin \left(- {225}^{\circ}\right)\right)$

= $\frac{1}{\sqrt{2}} \left(\cos \left({360}^{\circ} - {225}^{\circ}\right) + i \sin \left({360}^{\circ} - {225}^{\circ}\right)\right)$

= $\frac{1}{\sqrt{2}} \left(\cos {135}^{\circ} + i \sin {135}^{\circ}\right)$

= $\frac{1}{\sqrt{2}} \left(- \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right)$

= $- \frac{1}{2} + i \frac{1}{2}$