The correct option is B #160; r=4,θ =π/6 z=e^iπ /2 ( 2e^iπ /6 )^6/4. ( 2.e^ iπ /3 )^2=4.e^i ( π /2+π +2π /3 ) nbsp; ∴ arg z=13π/6=2π +π/6=π/6 ...

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Integration by Substitution

i √3+i 6/4 1-...

Question

$\frac{i {(\sqrt{3 + i})}^{6}}{4 {(1 - i \sqrt{3})}^{2}}$

r = 2, θ = \frac{π}{3}

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r = 2, θ = - \frac{π}{3}

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r = 4, θ = \frac{π}{6}

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r = 4, θ = - \frac{π}{6}

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Solution

The correct option is B $r = 4, θ = \frac{π}{6}$
$z = \frac{e^{i π / 2} {(2 e^{i π / 6})}^{6}}{4. {(2. e^{- i π / 3})}^{2}} = 4. e^{i (π / 2 + π + 2 π / 3)}$
$∴ a r g z = \frac{13 π}{6} = 2 π + \frac{π}{6} = \frac{π}{6}$
$∴ r = 4, θ = \frac{π}{6}$
Ans: C

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i(√3+i)64(1−i√3)2

The correct option is B r=4,θ=π6z=eiπ/2(2eiπ/6)64.(2.e−iπ/3)2=4.ei(π/2+π+2π/3) ∴argz=13π6=2π+π6=π6 ∴r=4,θ=π6Ans: C

$\frac{i {(\sqrt{3 + i})}^{6}}{4 {(1 - i \sqrt{3})}^{2}}$

The correct option is B $r = 4, θ = \frac{π}{6}$
$z = \frac{e^{i π / 2} {(2 e^{i π / 6})}^{6}}{4. {(2. e^{- i π / 3})}^{2}} = 4. e^{i (π / 2 + π + 2 π / 3)}$
$∴ a r g z = \frac{13 π}{6} = 2 π + \frac{π}{6} = \frac{π}{6}$
$∴ r = 4, θ = \frac{π}{6}$
Ans: C