Question Number 17391 by tawa tawa last updated on 05/Jul/17
$$\mathrm{write}\:\:\:\mathrm{z}\:=\:\left(\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{i}\right)^{\mathrm{3}} \:\:\mathrm{in}\:\mathrm{polar}\:\mathrm{form}. \\ $$
Answered by ajfour last updated on 05/Jul/17
$$\mathrm{z}=\mathrm{4}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{64}\left(\mathrm{e}^{\boldsymbol{\mathrm{i}}\pi/\mathrm{3}} \right)^{\mathrm{3}} \\ $$$$\:=\mathrm{64e}^{\boldsymbol{\mathrm{i}}\pi} \:. \\ $$
Commented by tawa tawa last updated on 05/Jul/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by ajfour last updated on 05/Jul/17
$$\mathrm{loss}\:\mathrm{of}\:\mathrm{sleep}\:\mathrm{sir}.\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by mrW1 last updated on 05/Jul/17
$$\mathrm{is}\:\mathrm{z}=−\mathrm{64}? \\ $$
Commented by b.e.h.i.8.3.4.17@gmail.com last updated on 06/Jul/17
$${yes}\:.{because}\:{of}\:{Euler}'{s}\:{rule}: \\ $$$$\boldsymbol{{e}}^{\boldsymbol{{i}\pi}} +\mathrm{1}=\mathrm{0}\:\:. \\ $$