calculate-i-1-49-cos-pi-40-i-sin-pi-40-10-

Question Number 56144 by gunawan last updated on 11/Mar/19

calculate {(i - 1)}^{49} {(\cos \frac{π}{40} + i \sin \frac{π}{40})}^{10}

Answered by Smail last updated on 11/Mar/19

{(i - 1)}^{49} = {[\sqrt{2} (\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2})]}^{49}

= \sqrt{2^{49}} {(e^{i \frac{π}{4}})}^{49} = 2^{24} \sqrt{2} e^{i \frac{49 π}{4}} = 2^{24} \sqrt{2} e^{i (\frac{π}{4} + 12 π)}

= 2^{24} \sqrt{2} e^{i \frac{π}{4}}

{(c o s \frac{π}{40} + i s i n \frac{π}{40})}^{10} = {(e^{i \frac{π}{40}})}^{10} = e^{i \frac{π}{4}}

{(i - 1)}^{49} {(c o s \frac{π}{40} + i s i n \frac{π}{40})}^{10} = (2^{24} \sqrt{2} e^{i \frac{π}{4}}) e^{i \frac{π}{4}}

= 2^{24} \sqrt{2} e^{i \frac{π}{2}}

= 2^{24} \sqrt{2} i

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