For any integer n- the argument of z -3- i 4 n -1-1- i -34 n isTA- 6B- -3C-D- 2 -2

Given: z = (√(3) + i)^4n + 1(1 i√(3))^4n ⇒ z= i^4n(√(3) + i)^4n + 1(i +√(3))^4n ⇒ z= √(3) + i tanθ = 1√(3) As the complex number lies in the first qu ...

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Standard XII

Mathematics

Geometrical Representation of a Complex Number

For any integ...

Question

For any integer $n,$ the argument of $z = \frac{(\sqrt{3} + i)^{4 n + 1}}{(1 - i \sqrt{3})^{4 n}}$ is

\frac{π}{6}

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\frac{π}{3}

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\frac{π}{2}

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\frac{2 π}{3}

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Solution

The correct option is A $\frac{π}{6}$
$z = \frac{(\sqrt{3} + i)^{4 n + 1}}{(1 - i \sqrt{3})^{4 n}} = \frac{i^{4 n} (\sqrt{3} + i)^{4 n + 1}}{(i + \sqrt{3})^{4 n}} = \sqrt{3} + i$
$\Rightarrow arg (z) = \frac{π}{6}$

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Geometrical Representation of a Complex Number

Standard XII Mathematics

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For any integer n, the argument of z=(√3+i)4n+1(1−i√3)4n is

The correct option is A π6z=(√3+i)4n+1(1−i√3)4n=i4n(√3+i)4n+1(i+√3)4n=√3+i ⇒arg(z)=π6

For any integer $n,$ the argument of $z = \frac{(\sqrt{3} + i)^{4 n + 1}}{(1 - i \sqrt{3})^{4 n}}$ is

The correct option is A $\frac{π}{6}$
$z = \frac{(\sqrt{3} + i)^{4 n + 1}}{(1 - i \sqrt{3})^{4 n}} = \frac{i^{4 n} (\sqrt{3} + i)^{4 n + 1}}{(i + \sqrt{3})^{4 n}} = \sqrt{3} + i$
$\Rightarrow arg (z) = \frac{π}{6}$