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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 but 1−i =(√2)((1/(√2)) −(i/(√2))) =(√2)e^(−i(π/4)) ⇒ (1−i)^6 =2^3 e^(−i((6π)/4)) =8 e^(−((i3π)/2)) ⇒ ∣z_1 z_2 ∣=8 and arg(z_1 z_2 ) ≡ −((3π)/2)≡(π/2)[2π]

$${z}_{\mathrm{1}} {z}_{\mathrm{2}} =\left(\mathrm{1}−{i}\right)^{\mathrm{6}} \:\:{but}\:\mathrm{1}−{i}\:=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:−\frac{{i}}{\sqrt{\mathrm{2}}}\right)\:=\sqrt{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\left(\mathrm{1}−{i}\right)^{\mathrm{6}} =\mathrm{2}^{\mathrm{3}} \:{e}^{−{i}\frac{\mathrm{6}\pi}{\mathrm{4}}} \:=\mathrm{8}\:{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{2}}} \:\:\Rightarrow\:\mid{z}_{\mathrm{1}} {z}_{\mathrm{2}} \mid=\mathrm{8}\:{and}\: \\ $$$${arg}\left({z}_{\mathrm{1}} {z}_{\mathrm{2}} \right)\:\equiv\:−\frac{\mathrm{3}\pi}{\mathrm{2}}\equiv\frac{\pi}{\mathrm{2}}\left[\mathrm{2}\pi\right] \\ $$$$ \\ $$

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 − 6i − 15 + 20i + 15 − 6i −1 z_1 z_2 = 8i Modulus = 8 Argument = lim_(x→0) (tan^(−1) ((8/x))) = (π/2)

$${z}_{\mathrm{1}} {z}_{\mathrm{2}} \:=\:\left(\mathrm{1}−{i}\right)^{\mathrm{13}} \left(\mathrm{1}−{i}\right)^{−\mathrm{7}} \\ $$$${z}_{\mathrm{1}} {z}_{\mathrm{2}} \:=\:\left(\mathrm{1}−{i}\right)^{\mathrm{6}} \\ $$$${z}_{\mathrm{1}} {z}_{\mathrm{2}} \:=\:\mathrm{1}\:+\:\mathrm{6}\left(−{i}\right)\:+\:\mathrm{15}\left(−{i}\right)^{\mathrm{2}} \:+\:\mathrm{20}\left(−{i}\right)^{\mathrm{3}} \:+\:\mathrm{15}\left(−{i}\right)^{\mathrm{4}} \:+\:\mathrm{6}\left(−{i}\right)^{\mathrm{5}} \:+\:\left(−{i}\right)^{\mathrm{6}} \\ $$$${z}_{\mathrm{1}} {z}_{\mathrm{2}} \:=\:\mathrm{1}\:+\:\mathrm{6}\left(−{i}\right)\:+\:\mathrm{15}\left(−\mathrm{1}\right)\:+\:\mathrm{20}\left({i}\right)\:+\:\mathrm{15}\left(\mathrm{1}\right)\:+\:\mathrm{6}\left(−{i}\right)\:+\:\left(−\mathrm{1}\right) \\ $$$${z}_{\mathrm{1}} {z}_{\mathrm{2}} \:=\:\mathrm{1}\:−\:\mathrm{6}{i}\:−\:\mathrm{15}\:+\:\mathrm{20}{i}\:+\:\mathrm{15}\:−\:\mathrm{6}{i}\:−\mathrm{1} \\ $$$${z}_{\mathrm{1}} {z}_{\mathrm{2}} \:=\:\mathrm{8}{i} \\ $$$$ \\ $$$${Modulus}\:=\:\mathrm{8} \\ $$$${Argument}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{8}}{{x}}\right)\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Nov/18

(1−i)^6 =(1−2i+i^2 )^3 =(−2i)^3 =−8×i^2 ×i=8i modulus∣0+8i∣ is(√(0^2 +8^2 )) =8 argument tan^(−1) ((8/0)) that is (π/2)

$$\left(\mathrm{1}−{i}\right)^{\mathrm{6}} =\left(\mathrm{1}−\mathrm{2}{i}+{i}^{\mathrm{2}} \right)^{\mathrm{3}} =\left(−\mathrm{2}{i}\right)^{\mathrm{3}} =−\mathrm{8}×{i}^{\mathrm{2}} ×{i}=\mathrm{8}{i} \\ $$$${modulus}\mid\mathrm{0}+\mathrm{8}{i}\mid\:\:{is}\sqrt{\mathrm{0}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }\:=\mathrm{8} \\ $$$${argument}\:\:{tan}^{−\mathrm{1}} \left(\frac{\mathrm{8}}{\mathrm{0}}\right)\:{that}\:{is}\:\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$

Answered by MJS last updated on 28/Nov/18

z_1 z_2 =(1−i)^6 z=1−i=re^(iθ) ; r=(√2)∧θ=((7π)/4) ⇒ z=(√2)e^(i((7π)/4)) z^6 =r^6 e^(6iθ) =8e^(i((21π)/2)) =8e^(i(π/2)) ⇒ r=8 ∧ θ=(π/2)

$${z}_{\mathrm{1}} {z}_{\mathrm{2}} =\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{6}} \\ $$$${z}=\mathrm{1}−\mathrm{i}={r}\mathrm{e}^{\mathrm{i}\theta} ;\:{r}=\sqrt{\mathrm{2}}\wedge\theta=\frac{\mathrm{7}\pi}{\mathrm{4}}\:\Rightarrow\:{z}=\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{i}\frac{\mathrm{7}\pi}{\mathrm{4}}} \\ $$$${z}^{\mathrm{6}} ={r}^{\mathrm{6}} \mathrm{e}^{\mathrm{6i}\theta} =\mathrm{8e}^{\mathrm{i}\frac{\mathrm{21}\pi}{\mathrm{2}}} =\mathrm{8e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \:\Rightarrow\:{r}=\mathrm{8}\:\wedge\:\theta=\frac{\pi}{\mathrm{2}} \\ $$

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