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Find the indicated power using De Moivre's Theorem. (Express your fully simplified answer in the form

a + bi.)

(1-i)^18

Apr 20, 2020

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Find the indicated power using De Moivre's Theorem. (Express your fully simplified answer in the form
$$a + bi$$.)

$$\left(1-i\right)^{18}$$

$$\begin{array}{|rcll|} \hline 1-i &=& r\Big(\cos(x) -+ i \sin(x) \Big) \\ && \boxed{ r = \sqrt{1^2+(-1)^2} \\ r=\sqrt{2} } \\ && \boxed{ x = \arctan\left(\dfrac{-1}{1}\right) \\x = \arctan(-1)\\x = -45^\circ } \\ 1-i &=& \sqrt{2}\Big(\cos(-45^\circ) + i \sin(-45^\circ) \Big) \\ 1-i &=& \sqrt{2}\Big(\cos(45^\circ) - i \sin(45^\circ) \Big) \\ \left(1-i\right)^{18}&=& \left(\sqrt{2}\right)^{18}\Big(\cos(18*45^\circ) - i \sin(18*45^\circ) \Big) \quad | \quad \text{De Moivre's Theorem} \\ \left(1-i\right)^{18}&=& 2^{\frac{18}{2}}\Big(\cos(810^\circ) - i \sin(810^\circ) \Big) \\ \left(1-i\right)^{18}&=& 2^{9}\Big(\cos(90^\circ) - i \sin(90^\circ) \Big) \\ \left(1-i\right)^{18}&=& 512\Big(0 - i*1 \Big) \\ \mathbf{\left(1-i\right)^{18}} &=& \mathbf{-512i} \\ \hline \end{array}$$

Apr 20, 2020
edited by heureka  Apr 21, 2020