Question Number 108930 by mohammad17 last updated on 20/Aug/20
Answered by ajfour last updated on 20/Aug/20
$${z}=\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{15}} }{\mathrm{128}}×\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{15}} }{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{15}} }\:=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{15}} \\ $$$$\:\:\:=\:\sqrt{\mathrm{2}}{e}^{\mathrm{15}{i}\pi/\mathrm{4}} \:=\:\sqrt{\mathrm{2}}{e}^{−{i}\pi/\mathrm{4}} \\ $$$$\bar {{z}}=\sqrt{\mathrm{2}}{e}^{{i}\pi/\mathrm{4}} \:=\:\mathrm{1}+{i}\:. \\ $$
Commented by mohammad17 last updated on 20/Aug/20
$${thank}\:{you}\:{sir} \\ $$
Answered by Dwaipayan Shikari last updated on 20/Aug/20
$$\frac{\left(\mathrm{1}+{i}\right)^{\frac{\mathrm{15}}{\mathrm{2}}.\mathrm{2}} }{\mathrm{128}}=\frac{\left(\mathrm{2}{i}\right)^{\frac{\mathrm{15}}{\mathrm{2}}} }{\mathrm{128}}=\mathrm{128}\sqrt{\mathrm{2}}.\frac{\left({i}\right)^{\mathrm{7}+\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{128}}=\sqrt{\mathrm{2}}.\left(−{i}\sqrt{{i}}\right)=\sqrt{\mathrm{2}}\left(−{i}\right).\left(\mathrm{1}+{i}\right).\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\mathrm{1}−{i}={z} \\ $$$$\overset{−} {{z}}=\mathrm{1}+{i} \\ $$
Commented by mohammad17 last updated on 20/Aug/20
$${thank}\:{you}\:{sir} \\ $$