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Question Number 146521 by mathdanisur last updated on 13/Jul/21

(i - 1)^(−100) = ?

$$\left(\boldsymbol{{i}}\:-\:\mathrm{1}\right)^{−\mathrm{100}} \:=\:? \\ $$

Answered by mathmax by abdo last updated on 13/Jul/21

(i−1)^(−100) =(−1+i)^(−100) =((√2)e^((i3π)/4) )^(−100) =((√2))^(−100) e^(−((i3π)/4)×100) =(1/2^(50) ) e^(−(3iπ)×25) =(1/2^(50) )e^(−75iπ) =(1/2^(50) )e^(−74iπ) e^(−iπ) =−(1/2^(50) )

$$\left(\mathrm{i}−\mathrm{1}\right)^{−\mathrm{100}} =\left(−\mathrm{1}+\mathrm{i}\right)^{−\mathrm{100}} \:=\left(\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i3}\pi}{\mathrm{4}}} \right)^{−\mathrm{100}} \:=\left(\sqrt{\mathrm{2}}\right)^{−\mathrm{100}} \:\mathrm{e}^{−\frac{\mathrm{i3}\pi}{\mathrm{4}}×\mathrm{100}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{50}} }\:\mathrm{e}^{−\left(\mathrm{3i}\pi\right)×\mathrm{25}} \:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{50}} }\mathrm{e}^{−\mathrm{75i}\pi} \:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{50}} }\mathrm{e}^{−\mathrm{74i}\pi} \:\mathrm{e}^{−\mathrm{i}\pi} \:=−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{50}} } \\ $$

Commented by mathdanisur last updated on 13/Jul/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

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