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

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

Answered by liberty last updated on 03/Jan/21

$${(1+\mathrm{i})}^{2020}={\left(\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}})\right)}^{2020}$$$$={\left(\sqrt{2}\right)}^{2020}{(\cos \left(\frac{\pi }{4}\right)+\mathrm{i}\sin \left(\frac{\pi }{4}\right))}^{2020}$$$$=2^{1010}(\cos \left(505\pi \right)+\mathrm{i}\sin \left(505\pi \right))$$$$=2^{1010}(-1+0)=-2^{1010}$$

Commented by JDamian last updated on 03/Jan/21

$$\mathbf{MJS}\mathbf{\_}\mathbf{new}\mathrm{is}\mathrm{right}-\cos \left(505\pi \right)is-1$$$$\cos \left(\mathrm{k}\pi \right)={(-1)}^k\mathrm{\forall }k\in \mathbf{Z}$$

Answered by MJS_new last updated on 03/Jan/21

$$1+\mathrm{i}=\sqrt{2}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}$$$${(1+\mathrm{i})}^{2020}={\left(\sqrt{2}\right)}^{2020}{\mathrm{e}}^{\mathrm{i}\frac{2020\pi }{4}}=2^{1010}{\mathrm{e}}^{505\pi \mathrm{i}}=-2^{1010}$$

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