Simplified form of -3- i 6-1- i 8A- 4 iB- 4-4 iC- -4D- 0

Given: (√(3) i)^6(1+i)^8 √(3) i =2(√(3)2 i2) nbsp;⇒√(3) i = 2(cosπ6 isinπ6) ⇒ ( √(3) i )^6 = nbsp; (2( cosπ6 isinπ6))^6 nbsp; ⇒ ( √(3) i )^6= 64(cos ...

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Standard XII

Mathematics

nth Root of a Complex Number

Simplified fo...

Question

Simplified form of $\frac{(\sqrt{3} - i)^{6}}{(1 + i)^{8}}$ is

4 i

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4 - 4 i

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- 4

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0

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Solution

The correct option is C $- 4$
$∵ \sqrt{3} - i = 2 (cos \frac{π}{6} - i sin \frac{π}{6})$
$∴ {(\sqrt{3} - i)}^{6} = 64 (cos π - i sin π) = - 64$
Similarly, $1 + i = \sqrt{2} (cos \frac{π}{4} + i sin \frac{π}{4})$
$∴ (1 + i)^{8} = 16 (cos 2 π + i sin 2 π) = 16$
So, $\frac{(\sqrt{3} - i)^{6}}{(1 + i)^{8}} = - \frac{64}{16}$
$= - 4$

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nth Root of a Complex Number

Standard XII Mathematics

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Simplified form of (√3−i)6(1+i)8 is

The correct option is C −4∵√3−i=2(cosπ6−isinπ6) ∴(√3−i)6=64(cosπ−isinπ)=−64 Similarly, 1+i=√2(cosπ4+isinπ4) ∴(1+i)8=16(cos2π+isin2π)=16 So, (√3−i)6(1+i)8=−6416 =−4

Simplified form of $\frac{(\sqrt{3} - i)^{6}}{(1 + i)^{8}}$ is