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Question Number 48740 by peter frank last updated on 28/Nov/18

Commented by Abdo msup. last updated on 28/Nov/18

$z_{1} z_{2} = {(1 - i)}^{6} b u t 1 - i = \sqrt{2} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \sqrt{2} e^{- i \frac{π}{4}} \Rightarrow$ ${(1 - i)}^{6} = 2^{3} e^{- i \frac{6 π}{4}} = 8 e^{- \frac{i 3 π}{2}} \Rightarrow ∣ z_{1} z_{2} ∣= 8 a n d$ $a r g (z_{1} z_{2}) \equiv - \frac{3 π}{2} \equiv \frac{π}{2} [2 π]$
	Answered by Abdulhafeez Abu qatada last updated on 28/Nov/18

	$z_{1} z_{2} = {(1 - i)}^{13} {(1 - i)}^{- 7}$ $z_{1} z_{2} = {(1 - i)}^{6}$ $z_{1} z_{2} = 1 + 6 (- i) + 15 {(- i)}^{2} + 20 {(- i)}^{3} + 15 {(- i)}^{4} + 6 {(- i)}^{5} + {(- i)}^{6}$ $z_{1} z_{2} = 1 + 6 (- i) + 15 (- 1) + 20 (i) + 15 (1) + 6 (- i) + (- 1)$ $z_{1} z_{2} = 1 - 6 i - 15 + 20 i + 15 - 6 i - 1$ $z_{1} z_{2} = 8 i$ $M o d u l u s = 8$ $A r g u m e n t = \lim_{x \to 0} (\tan^{- 1} (\frac{8}{x})) = \frac{π}{2}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 28/Nov/18

	${(1 - i)}^{6} = {(1 - 2 i + i^{2})}^{3} = {(- 2 i)}^{3} = - 8 \times i^{2} \times i = 8 i$ $m o d u l u s ∣ 0 + 8 i ∣ i s \sqrt{0^{2} + 8^{2}} = 8$ $a r g u m e n t {t a n}^{- 1} (\frac{8}{0}) t h a t i s \frac{π}{2}$
	Answered by MJS last updated on 28/Nov/18

	$z_{1} z_{2} = {(1 - i)}^{6}$ $z = 1 - i = r e^{i θ}; r = \sqrt{2} \land θ = \frac{7 π}{4} \Rightarrow z = \sqrt{2} e^{i \frac{7 π}{4}}$ $z^{6} = r^{6} e^{6 i θ} = {8 e}^{i \frac{21 π}{2}} = {8 e}^{i \frac{π}{2}} \Rightarrow r = 8 \land θ = \frac{π}{2}$
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