If z=x+iy, x^2+y^2=1 and z_1=ze^{itheta}, the | vedclass

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

Question

(D) Given $z=x+iy$ and $x^2+y^2=1$, we can write $z=e^{i\phi}$ where $\phi = 	an^{-1}(\frac{y}{x})$.$z_1 = ze^{i	heta} = e^{i\phi}e^{i	heta} = e^{i

Vedclass · Accepted Answer

$i 	an (n(	heta+	an^{-1} \frac{y}{x}))$

Vedclass · Answer

(D) Given $z=x+iy$ and $x^2+y^2=1$, we can write $z=e^{i\phi}$ where $\phi = 	an^{-1}(\frac{y}{x})$.$z_1 = ze^{i	heta} = e^{i\phi}e^{i	heta} = e^{i(\phi+	heta)}$.Then $z_1^{2n} = e^{i2n(\phi+	heta)}$.Consider the expression $\frac{z_1^{2n}-1}{z_1^{2n}+1} = \frac{e^{i2n(\phi+	heta)}-1}{e^{i2n(\phi+	heta)}+1}$.Multiply numerator and denominator by $e^{-in(\phi+	heta)}$:$= \frac{e^{in(\phi+	heta)} - e^{-in(\phi+	heta)}}{e^{in(\phi+	heta)} + e^{-in(\phi+	heta)}} = \frac{2i \sin(n(\phi+	heta))}{2 \cos(n(\phi+	heta))}$.$= i 	an(n(\phi+	heta)) = i 	an(n(	heta+	an^{-1}(\frac{y}{x})))$.

If $z=x+iy$, $x^2+y^2=1$ and $z_1=ze^{i\theta}$, then $\frac{z_1^{2n}-1}{z_1^{2n}+1}=$

Similar Questions

If $\omega$ is a complex cube root of unity,then $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors is equal to:

If $\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}(x+i y)$ where $x$ and $y$ are real, then the ordered pair $(x, y)$ is

The real part of $z$ that satisfies $i z^4+1=0$ is

$(-1+i \sqrt{3})^{60} = ?$

If the complex cube roots of $(-i)$ are $\alpha, \beta, \gamma$,then $\alpha^2+\beta^2+\gamma^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module