write-1-i-23-3-i-13-in-re-i-

Question Number 132950 by mohammad17 last updated on 17/Feb/21

$${write}\:\frac{\left(\mathrm{1}−{i}\right)^{\mathrm{23}} }{\left(\sqrt{\mathrm{3}}−{i}\right)^{\mathrm{13}} }\:{in}\left(\:{re}^{{i}\theta} \right) \\ $$

Answered by metamorfose last updated on 17/Feb/21

z=(((1−i)^(23) )/( ((√3)−i)^(13) ))=(1/(2(√2))) e^(i((5π)/(12)))

$${z}=\frac{\left(\mathrm{1}−{i}\right)^{\mathrm{23}} }{\:\left(\sqrt{\mathrm{3}}−{i}\right)^{\mathrm{13}} }=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{12}}} \\ $$

Commented by metamorfose last updated on 17/Feb/21

ok.. z=(((1−i)^(23) )/(((√3)−i)^(13) ))=((((√2)e^(−i(π/4)) )^(23) )/((2e^(−i(π/6)) )^(13) ))=((((√2))^(23) e^(−i((23π)/4)) )/(2^(13) e^(−i((13π)/6)) )) =((((√2))^(23) )/(((√2))^(26) ))e^(−i(((23π)/4)−((13π)/6))) =(1/( ((√2))^3 ))e^(−i(π/2)(((23)/2)−((13)/3))) =(1/(2(√2)))e^(−i((43π)/(12))) =(1/(2(√2)))e^(−i(((48π)/(12))−((5π)/(12)))) =(1/(2(√2)))e^(−i(4π−((5π)/(12)))) =(1/(2(√2)))e^(i((5π)/(12)))

$${ok}.. \\ $$$${z}=\frac{\left(\mathrm{1}−{i}\right)^{\mathrm{23}} }{\left(\sqrt{\mathrm{3}}−{i}\right)^{\mathrm{13}} }=\frac{\left(\sqrt{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{23}} }{\left(\mathrm{2}{e}^{−{i}\frac{\pi}{\mathrm{6}}} \right)^{\mathrm{13}} }=\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{23}} {e}^{−{i}\frac{\mathrm{23}\pi}{\mathrm{4}}} }{\mathrm{2}^{\mathrm{13}} {e}^{−{i}\frac{\mathrm{13}\pi}{\mathrm{6}}} } \\ $$$$\:\:\:\:=\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{23}} }{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{26}} }{e}^{−{i}\left(\frac{\mathrm{23}\pi}{\mathrm{4}}−\frac{\mathrm{13}\pi}{\mathrm{6}}\right)} =\frac{\mathrm{1}}{\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{3}} }{e}^{−{i}\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{23}}{\mathrm{2}}−\frac{\mathrm{13}}{\mathrm{3}}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{e}^{−{i}\frac{\mathrm{43}\pi}{\mathrm{12}}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{e}^{−{i}\left(\frac{\mathrm{48}\pi}{\mathrm{12}}−\frac{\mathrm{5}\pi}{\mathrm{12}}\right)} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{e}^{−{i}\left(\mathrm{4}\pi−\frac{\mathrm{5}\pi}{\mathrm{12}}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{12}}} \\ $$

Commented by mohammad17 last updated on 17/Feb/21