For any integer n- the argument of z - -3 - i4n - 11 - i-34n is

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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{π}{6}$
$z = \frac{(\sqrt{3} + i)^{4 n + 1}}{(1 - i \sqrt{3})^{4 n}} = \frac{i^{4 n} (\sqrt{3} + i)^{4 n + 1}}{(i + \sqrt{3})^{4 n}} = \sqrt{3} + i$
$tan θ = \frac{1}{\sqrt{3}}$
As the complex number lies in the first quadrant, so
$arg (z) = θ = \frac{π}{6}$

Suggest Corrections

Similar questions

Q. For any integer

n

, the argument of

z = \frac{{(\sqrt{3} + i)}^{4 n + 1}}{{(1 - i \sqrt{3})}^{4 n}}

Q. For any integer

n,

the argument of

z = \frac{(\sqrt{3} + i)^{4 n + 1}}{(1 - i \sqrt{3})^{4 n}}

Q. For n an integer, the argument of

Z = \frac{(\sqrt{3} + i)^{4 n + 1}}{(1 - i \sqrt{3})^{4 n}}

Q. For any integer

n

, find the modulus of

z = \frac{(\sqrt{3} + i)^{4 n + 1}}{(1 - i \sqrt{3})^{4 n}}

If n is any positive integer, write the value of $\frac{i^{4 n + 1} - i^{4 n - 1}}{2}$

Join BYJU'S Learning Program

Explore more

Geometrical Representation of a Complex Number

Standard XII Mathematics

Join BYJU'S Learning Program

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 tanθ=1√3 As the complex number lies in the first quadrant, so arg(z)=θ=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 $\frac{π}{6}$
$z = \frac{(\sqrt{3} + i)^{4 n + 1}}{(1 - i \sqrt{3})^{4 n}} = \frac{i^{4 n} (\sqrt{3} + i)^{4 n + 1}}{(i + \sqrt{3})^{4 n}} = \sqrt{3} + i$
$tan θ = \frac{1}{\sqrt{3}}$
As the complex number lies in the first quadrant, so
$arg (z) = θ = \frac{π}{6}$