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Question Number 133346 by bramlexs22 last updated on 21/Feb/21
 Find modulus and argumen of    z = (((1−i)^4 ((√3)+i)^7 )/((1+i(√2))^8 (−1−i(√3))^(12) ))
$$\:\mathrm{Find}\:\mathrm{modulus}\:\mathrm{and}\:\mathrm{argumen}\:\mathrm{of}\: \\ $$$$\:\mathrm{z}\:=\:\frac{\left(\mathrm{1}−{i}\right)^{\mathrm{4}} \left(\sqrt{\mathrm{3}}+{i}\right)^{\mathrm{7}} }{\left(\mathrm{1}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{8}} \left(−\mathrm{1}−{i}\sqrt{\mathrm{3}}\right)^{\mathrm{12}} } \\ $$
Answered by mathmax by abdo last updated on 21/Feb/21
z_1 =(1−i)^4 ((√3)+i)^7  and z_2 =(1+i(√2))^8 (−1−i(√3))^(12)  ⇒  z =(z_1 /z_2 ) ⇒ ∣z∣=((∣z_1 ∣)/(∣z_2 ∣)) and arg(z)∼arg(z_1 )−arg(z_2 )[2π]  ∣1−i∣ =(√2) ⇒1−i =(√2)((1/( (√2)))−(i/( (√2))))=(√2)e^(−((iπ)/4))  ⇒(1−i)^4  =((√2))^4  e^(−iπ)   ∣(√3)+i∣=2 ⇒(√3)+i =2(((√3)/2)+(1/2)i) =2e^((iπ)/6)  ⇒((√3)+i)^7  =2^7  e^((i7π)/6)  ⇒  z_1 =2^9  e^(−iπ) .e^(i((7π)/6))  =2^9  e^(i(((7π)/6)−π))  =2^9  e^((iπ)/6)   ∣1+i(√2)∣=(√3) ⇒1+i(√2)=(√3)((1/( (√3)))+i((√2)/( (√3)))) =(√3)e^(iarctan((√2)))   ⇒(1+i(√2))^8  =((√3))^(8 ) .e^(8iarctan((√2)))   ∣−1−i(√3)∣ =2 ⇒−1−i(√3)=2(−(1/2)−i((√3)/2)) =2e^(−((i4π)/3))  ⇒  (−1−i(√3))^(12)  =2^(12) .e^(−i16π)  ⇒z_2 =((√3))^8  .e^(8iarctan((√2))) .2^(12)  .e^(−16iπ)  ⇒  z =((2^9  e^((iπ)/6) )/(2^(12) .((√3))^8  .e^(8iarctan((√2))) )) =(1/(8.((√3))^8 ))e^(i((π/6)−8arctan((√2))))  ⇒  ∣z∣=(1/(8.((√3))^8 )) and arg(z)∼ (π/6)−8arctan((√2)) [2π]
$$\mathrm{z}_{\mathrm{1}} =\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{4}} \left(\sqrt{\mathrm{3}}+\mathrm{i}\right)^{\mathrm{7}} \:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\left(\mathrm{1}+\mathrm{i}\sqrt{\mathrm{2}}\right)^{\mathrm{8}} \left(−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{12}} \:\Rightarrow \\ $$$$\mathrm{z}\:=\frac{\mathrm{z}_{\mathrm{1}} }{\mathrm{z}_{\mathrm{2}} }\:\Rightarrow\:\mid\mathrm{z}\mid=\frac{\mid\mathrm{z}_{\mathrm{1}} \mid}{\mid\mathrm{z}_{\mathrm{2}} \mid}\:\mathrm{and}\:\mathrm{arg}\left(\mathrm{z}\right)\sim\mathrm{arg}\left(\mathrm{z}_{\mathrm{1}} \right)−\mathrm{arg}\left(\mathrm{z}_{\mathrm{2}} \right)\left[\mathrm{2}\pi\right] \\ $$$$\mid\mathrm{1}−\mathrm{i}\mid\:=\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{1}−\mathrm{i}\:=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)=\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\Rightarrow\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{4}} \:=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{4}} \:\mathrm{e}^{−\mathrm{i}\pi} \\ $$$$\mid\sqrt{\mathrm{3}}+\mathrm{i}\mid=\mathrm{2}\:\Rightarrow\sqrt{\mathrm{3}}+\mathrm{i}\:=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}\right)\:=\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{6}}} \:\Rightarrow\left(\sqrt{\mathrm{3}}+\mathrm{i}\right)^{\mathrm{7}} \:=\mathrm{2}^{\mathrm{7}} \:\mathrm{e}^{\frac{\mathrm{i7}\pi}{\mathrm{6}}} \:\Rightarrow \\ $$$$\mathrm{z}_{\mathrm{1}} =\mathrm{2}^{\mathrm{9}} \:\mathrm{e}^{−\mathrm{i}\pi} .\mathrm{e}^{\mathrm{i}\frac{\mathrm{7}\pi}{\mathrm{6}}} \:=\mathrm{2}^{\mathrm{9}} \:\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{7}\pi}{\mathrm{6}}−\pi\right)} \:=\mathrm{2}^{\mathrm{9}} \:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{6}}} \\ $$$$\mid\mathrm{1}+\mathrm{i}\sqrt{\mathrm{2}}\mid=\sqrt{\mathrm{3}}\:\Rightarrow\mathrm{1}+\mathrm{i}\sqrt{\mathrm{2}}=\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\mathrm{i}\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\:=\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{iarctan}\left(\sqrt{\mathrm{2}}\right)} \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{i}\sqrt{\mathrm{2}}\right)^{\mathrm{8}} \:=\left(\sqrt{\mathrm{3}}\right)^{\mathrm{8}\:} .\mathrm{e}^{\mathrm{8iarctan}\left(\sqrt{\mathrm{2}}\right)} \\ $$$$\mid−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}\mid\:=\mathrm{2}\:\Rightarrow−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:=\mathrm{2e}^{−\frac{\mathrm{i4}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$$\left(−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{12}} \:=\mathrm{2}^{\mathrm{12}} .\mathrm{e}^{−\mathrm{i16}\pi} \:\Rightarrow\mathrm{z}_{\mathrm{2}} =\left(\sqrt{\mathrm{3}}\right)^{\mathrm{8}} \:.\mathrm{e}^{\mathrm{8iarctan}\left(\sqrt{\mathrm{2}}\right)} .\mathrm{2}^{\mathrm{12}} \:.\mathrm{e}^{−\mathrm{16i}\pi} \:\Rightarrow \\ $$$$\mathrm{z}\:=\frac{\mathrm{2}^{\mathrm{9}} \:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{6}}} }{\mathrm{2}^{\mathrm{12}} .\left(\sqrt{\mathrm{3}}\right)^{\mathrm{8}} \:.\mathrm{e}^{\mathrm{8iarctan}\left(\sqrt{\mathrm{2}}\right)} }\:=\frac{\mathrm{1}}{\mathrm{8}.\left(\sqrt{\mathrm{3}}\right)^{\mathrm{8}} }\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{\mathrm{6}}−\mathrm{8arctan}\left(\sqrt{\mathrm{2}}\right)\right)} \:\Rightarrow \\ $$$$\mid\mathrm{z}\mid=\frac{\mathrm{1}}{\mathrm{8}.\left(\sqrt{\mathrm{3}}\right)^{\mathrm{8}} }\:\mathrm{and}\:\mathrm{arg}\left(\mathrm{z}\right)\sim\:\frac{\pi}{\mathrm{6}}−\mathrm{8arctan}\left(\sqrt{\mathrm{2}}\right)\:\left[\mathrm{2}\pi\right] \\ $$

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