1-i-2020-

Question Number 127979 by bobhans last updated on 03/Jan/21

{(1 + i)}^{2020} = ?

Answered by liberty last updated on 03/Jan/21

{(1 + i)}^{2020} = {(\sqrt{2} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}))}^{2020}

= {(\sqrt{2})}^{2020} {(\cos (\frac{π}{4}) + i \sin (\frac{π}{4}))}^{2020}

= 2^{1010} (\cos (505 π) + i \sin (505 π))

= 2^{1010} (- 1 + 0) = - 2^{1010}

Commented by JDamian last updated on 03/Jan/21

MJS_new is right - \cos (505 π) i s - 1

\cos (k π) = {(- 1)}^{k} \forall k \in Z

Answered by MJS_new last updated on 03/Jan/21

1 + i = \sqrt{2} e^{i \frac{π}{4}}

{(1 + i)}^{2020} = {(\sqrt{2})}^{2020} e^{i \frac{2020 π}{4}} = 2^{1010} e^{505 π i} = - 2^{1010}

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