obtain-the-modulus-and-arguement-of-1-i-4-2-2-3i-3-

Question Number 99623 by hardylanes last updated on 22/Jun/20

o b t a i n t h e m o d u l u s a n d a r g u e m e n t o f

\frac{{(1 - i)}^{4}}{(2 + 2 \sqrt{{3 i)}^{3}}}

Commented by Dwaipayan Shikari last updated on 22/Jun/20

= \frac{{(1 - i)}^{2.2}}{4^{3} {(\frac{1}{2} + \frac{\sqrt{3}}{2} i)}^{3}} = \frac{{(- 2 i)}^{2}}{64 e^{\frac{π i}{3} . 3}} = \frac{- 4}{64 e^{π i}} = \frac{- 4}{- 64} = \frac{1}{16} + 0 . i = z

m o d (z) i s \frac{1}{16}

a r g (z) i s 0 [A s α = \tan^{- 1} (\frac{0}{\frac{1}{16}}) = 0] {A n d e^{π i} = - 1}

Answered by Rio Michael last updated on 22/Jun/20

let z = \frac{{(1 - i)}^{4}}{{(2 + 2 \sqrt{3} i)}^{3}} = \frac{{(\sqrt{2} e^{i \frac{7 π}{4}})}^{4}}{{(4 e^{i \frac{π}{3}})}^{3}} = \frac{4 e^{7 π i}}{4^{3} e^{π i}}

\Rightarrow ∣ z ∣ = \frac{4}{4^{3}} = \frac{1}{4^{2}} = \frac{1}{16}

\arg z = 7 π - π = 6 π = 3 (2 π)

aco - terminal angle = 0

\Rightarrow \arg z = 0

Commented by mathmax by abdo last updated on 23/Jun/20

z = \frac{{(1 - i)}^{4}}{{(2 + 2 \sqrt{3} i)}^{3}} we have 1 - i = \sqrt{2} e^{- \frac{i π}{4}} \Rightarrow {(1 - i)}^{4} = 4 e^{- i π}

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

z = \frac{1}{16} e^{- i π - \frac{4 i π}{3}} = \frac{1}{16} e^{- i (π + \frac{4 π}{3})} = \frac{1}{16} e^{- i (\frac{7 π}{3})} = \frac{1}{16} e^{- i (2 π + \frac{π}{3})} = \frac{1}{16} e^{- \frac{i π}{3}} \Rightarrow

∣ z ∣ = \frac{1}{16} and argz \equiv - \frac{π}{3} [2 π]