begin{aligned} & text{If } z=e^{i theta} text | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $z=e^{i 	heta}$,we have $\cos n 	heta = \frac{z^n+z^{-n}}{2}$ and $\sin n 	heta = \frac{z^n-z^{-n}}{2i}$.Substituting these into the expr

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given $z=e^{i 	heta}$,we have $\cos n 	heta = \frac{z^n+z^{-n}}{2}$ and $\sin n 	heta = \frac{z^n-z^{-n}}{2i}$.Substituting these into the expression:$\frac{3(\frac{z^3+z^{-3}}{2})+2(\frac{z^2+z^{-2}}{2})+5(\frac{z^5+z^{-5}}{2})}{3(\frac{z^3-z^{-3}}{2i})+2(\frac{z^2-z^{-2}}{2i})+5(\frac{z^5-z^{-5}}{2i})} = i \frac{3z^3+2z^2+5z^5+3z^{-3}+2z^{-2}+5z^{-5}}{3z^3+2z^2+5z^5-3z^{-3}-2z^{-2}-5z^{-5}}$$= i \frac{5z^{10}+3z^8+2z^7+2z^3+3z^2+5}{5z^{10}+3z^8+2z^7-2z^3-3z^2-5} = i \frac{\sum_{r=0}^{10} a_r z^r}{\sum_{r=0}^{10} b_r z^r}$.Comparing the coefficients,we find the sum of coefficients $\sum (a_r+b_r) = 2+3+5 = 10$.Therefore,$\frac{\sum_{r=0}^{10} (a_r+b_r)}{10} = \frac{10}{10} = 1$.

Similar Questions

If $(x-iy)^{1/3} = a-ib$,then the value of $\frac{x}{a} + \frac{y}{b}$ is

Which of the following are correct for any two complex numbers $z_1$ and $z_2$?

For all complex numbers $z$ of the form $1 + i\alpha$,where $\alpha \in R$,if $z^2 = x + iy$,then which of the following relations holds?

The modulus-amplitude form of $\frac{(1-i)^3(2-i)}{(2+i)(1+i)}$ is

The value of $\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module