Find-modulus-and-argumen-of-z-1-i-4-3-i-7-1-i-2-8-1-i-3-12-

Question Number 133346 by bramlexs22 last updated on 21/Feb/21

Find modulus and argumen of

z = \frac{{(1 - i)}^{4} {(\sqrt{3} + i)}^{7}}{{(1 + i \sqrt{2})}^{8} {(- 1 - i \sqrt{3})}^{12}}

Answered by mathmax by abdo last updated on 21/Feb/21

z_{1} = {(1 - i)}^{4} {(\sqrt{3} + i)}^{7} and z_{2} = {(1 + i \sqrt{2})}^{8} {(- 1 - i \sqrt{3})}^{12} \Rightarrow

z = \frac{z_{1}}{z_{2}} \Rightarrow ∣ z ∣= \frac{∣ z_{1} ∣}{∣ z_{2} ∣} and \arg (z) \sim \arg (z_{1}) - \arg (z_{2}) [2 π]

∣ 1 - i ∣ = \sqrt{2} \Rightarrow 1 - i = \sqrt{2} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \sqrt{2} e^{- \frac{i π}{4}} \Rightarrow {(1 - i)}^{4} = {(\sqrt{2})}^{4} e^{- i π}

∣ \sqrt{3} + i ∣= 2 \Rightarrow \sqrt{3} + i = 2 (\frac{\sqrt{3}}{2} + \frac{1}{2} i) = {2 e}^{\frac{i π}{6}} \Rightarrow {(\sqrt{3} + i)}^{7} = 2^{7} e^{\frac{i 7 π}{6}} \Rightarrow

z_{1} = 2^{9} e^{- i π} . e^{i \frac{7 π}{6}} = 2^{9} e^{i (\frac{7 π}{6} - π)} = 2^{9} e^{\frac{i π}{6}}

∣ 1 + i \sqrt{2} ∣= \sqrt{3} \Rightarrow 1 + i \sqrt{2} = \sqrt{3} (\frac{1}{\sqrt{3}} + i \frac{\sqrt{2}}{\sqrt{3}}) = \sqrt{3} e^{iarctan (\sqrt{2})}

\Rightarrow {(1 + i \sqrt{2})}^{8} = {(\sqrt{3})}^{8} . e^{8 iarctan (\sqrt{2})}

∣ - 1 - i \sqrt{3} ∣ = 2 \Rightarrow - 1 - i \sqrt{3} = 2 (- \frac{1}{2} - i \frac{\sqrt{3}}{2}) = {2 e}^{- \frac{i 4 π}{3}} \Rightarrow

{(- 1 - i \sqrt{3})}^{12} = 2^{12} . e^{- i 16 π} \Rightarrow z_{2} = {(\sqrt{3})}^{8} . e^{8 iarctan (\sqrt{2})} . 2^{12} . e^{- 16 i π} \Rightarrow

z = \frac{2^{9} e^{\frac{i π}{6}}}{2^{12} . {(\sqrt{3})}^{8} . e^{8 iarctan (\sqrt{2})}} = \frac{1}{8 . {(\sqrt{3})}^{8}} e^{i (\frac{π}{6} - 8 \arctan (\sqrt{2}))} \Rightarrow

∣ z ∣= \frac{1}{8 . {(\sqrt{3})}^{8}} and \arg (z) \sim \frac{π}{6} - 8 \arctan (\sqrt{2}) [2 π]