1-i-2020-

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

$$\:\:\left(\mathrm{1}+{i}\right)^{\mathrm{2020}} \:=? \\ $$

Answered by liberty last updated on 03/Jan/21

(1+i)^(2020) = ((√2) ((1/( (√2) )) + (i/( (√2)))))^(2020) = ((√2))^(2020) (cos ((π/4))+i sin ((π/4)))^(2020) = 2^(1010) (cos (505π) + i sin (505π)) = 2^(1010) (−1+0) =− 2^(1010)

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

Commented by JDamian last updated on 03/Jan/21

MJS_new is right − cos(505π) is −1 cos(kπ)=(−1)^k ∀k∈Z

$$\boldsymbol{\mathrm{MJS\_new}}\:\mathrm{is}\:\mathrm{right}\:−\:\mathrm{cos}\left(\mathrm{505}\pi\right)\:{is}\:−\mathrm{1} \\ $$$$\mathrm{cos}\left(\mathrm{k}\pi\right)=\left(−\mathrm{1}\right)^{{k}} \:\:\:\forall{k}\in\boldsymbol{\mathrm{Z}} \\ $$

Answered by MJS_new last updated on 03/Jan/21

1+i=(√2)e^(i(π/4)) (1+i)^(2020) =((√2))^(2020) e^(i((2020π)/4)) =2^(1010) e^(505πi) =−2^(1010)

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

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