Find-the-pricipal-value-of-1-i-1-i-

Question Number 19936 by ajfour last updated on 18/Aug/17

$${Find}\:{the}\:{pricipal}\:{value}\:{of}\: \\ $$$$\:\:\:\:\:\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} \:. \\ $$

Commented by prof Abdo imad last updated on 22/Jun/18

(1−i)^(1+i) =((√2) e^(−((iπ)/4)) )^(1+i) =((√2))^(1+i) e^(−((iπ)/4)(1+i)) =(√2) ((√2))^i e^(−((iπ)/4) +(π/4)) =(√2) e^(iln((√2))−((iπ)/4)) e^(π/4) =(√2) e^(π/4) e^(i{ ln((√)2)−(π/4)}) =(√2) e^(π/4) { cos(ln((√2))−(π/4)) +i sin(ln(2)−(π/4))}.

$$\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} =\left(\sqrt{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{1}+{i}} \\ $$$$=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{1}+{i}} \:{e}^{−\frac{{i}\pi}{\mathrm{4}}\left(\mathrm{1}+{i}\right)} \\ $$$$=\sqrt{\mathrm{2}}\:\left(\sqrt{\mathrm{2}}\right)^{{i}} \:{e}^{−\frac{{i}\pi}{\mathrm{4}}\:+\frac{\pi}{\mathrm{4}}} \\ $$$$=\sqrt{\mathrm{2}}\:{e}^{{iln}\left(\sqrt{\mathrm{2}}\right)−\frac{{i}\pi}{\mathrm{4}}} \:{e}^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\sqrt{\mathrm{2}}\:{e}^{\frac{\pi}{\mathrm{4}}} \:{e}^{{i}\left\{\:{ln}\left(\sqrt{}\mathrm{2}\right)−\frac{\pi}{\mathrm{4}}\right\}} \\ $$$$=\sqrt{\mathrm{2}}\:{e}^{\frac{\pi}{\mathrm{4}}} \left\{\:{cos}\left({ln}\left(\sqrt{\mathrm{2}}\right)−\frac{\pi}{\mathrm{4}}\right)\:+{i}\:{sin}\left({ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}}\right)\right\}. \\ $$

Answered by sma3l2996 last updated on 18/Aug/17

(1−i)^(1+i) =[(√2)(((√2)/2)−i((√2)/2))]^(1+i) =((√2)e^(−i(π/4)) )^(1+i) =2^((1+i)/2) ×e^((π/4)(1−i)) =2^(1/2) ×2^(i/2) ×e^(π/4) ×e^(−i(π/4)) =(√2)×e^(i((ln2)/2)) ×e^(π/4) ×e^(−iπ/4) (1−i)^(1+i) =(√2)e^(π/4) ×e^(i(((2ln2−π)/4)))

$$\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} =\left[\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right]^{\mathrm{1}+{i}} =\left(\sqrt{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{1}+{i}} \\ $$$$=\mathrm{2}^{\frac{\mathrm{1}+{i}}{\mathrm{2}}} ×{e}^{\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−{i}\right)} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{2}^{\frac{{i}}{\mathrm{2}}} ×{e}^{\frac{\pi}{\mathrm{4}}} ×{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$$=\sqrt{\mathrm{2}}×{e}^{{i}\frac{{ln}\mathrm{2}}{\mathrm{2}}} ×{e}^{\pi/\mathrm{4}} ×{e}^{−{i}\pi/\mathrm{4}} \\ $$$$\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} =\sqrt{\mathrm{2}}{e}^{\pi/\mathrm{4}} ×{e}^{{i}\left(\frac{\mathrm{2}{ln}\mathrm{2}−\pi}{\mathrm{4}}\right)} \\ $$

Commented by ajfour last updated on 18/Aug/17

$${thanks}\:{sir},\:{correct}\:{answer}. \\ $$

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