Solveeit Logo

Question

Mathematics Question on Logarithmic Differentiation

The general value of the real angle θ\theta, which satisfies the equation, (cosθ\theta + i sinθ\theta) (cos2θ\theta + i sin2θ\theta)........ (cosnθ\theta + i sinnθ\theta) = 1 is given by, (assuming k is an integer)

A

2kπn+2\frac{2k\pi}{n+2}

B

4kπn(n+1)\frac{4k\pi}{n\left(n+1\right)}

C

4kπn+1\frac{4k\pi}{n+1}

D

6kπn(n+1)\frac{6k\pi}{n\left(n+1\right)}

Answer

4kπn(n+1)\frac{4k\pi}{n\left(n+1\right)}

Explanation

Solution

The correct option is(C): 4kπn(n+1)\frac{4k\pi}{n\left(n+1\right)}.

eiθei(2θ)ei(3θ)ei(nθ)=1ei(θ(n(n+1))2)=ei2kπθ=4kπn(n+1)e^{i\theta}\cdot e^{i\left(2\theta\right)}\cdot e^{i\left(3\theta\right)}\ldots e^{i\left(n\theta\right)} = 1\,\Rightarrow\,e^{i\left(\theta \frac{\left(n\left(n+1\right)\right)}{2}\right)} = e^{i2k\,\pi\,\Rightarrow\,\theta = \frac{4k\pi}{n\left(n+1\right)}}