The maximum distance from the origin of coord | vedclass

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

Question

(C) Let $z = r(\cos 	heta + i \sin 	heta)$.Then $\left| z + \frac{1}{z} ight| = a \implies \left| z + \frac{1}{z} ight|^2 = a^2$.Expanding this,

Vedclass · Accepted Answer

$\frac{1}{2}(\sqrt{a^2 + 4} + a)$

Vedclass · Answer

(C) Let $z = r(\cos 	heta + i \sin 	heta)$.Then $\left| z + \frac{1}{z} ight| = a \implies \left| z + \frac{1}{z} ight|^2 = a^2$.Expanding this,we get $\left| r(\cos 	heta + i \sin 	heta) + \frac{1}{r}(\cos 	heta - i \sin 	heta) ight|^2 = a^2$.This simplifies to $\left( r + \frac{1}{r} ight)^2 \cos^2 	heta + \left( r - \frac{1}{r} ight)^2 \sin^2 	heta = a^2$.Using $\cos^2 	heta = 1 - \sin^2 	heta$,we get $r^2 + \frac{1}{r^2} + 2 \cos 2	heta = a^2$.To maximize $r$,we need to minimize $\cos 2	heta$. The minimum value of $\cos 2	heta$ is $-1$ (at $	heta = \frac{\pi}{2}$).Substituting $\cos 2	heta = -1$,we get $r^2 + \frac{1}{r^2} - 2 = a^2$.This is $(r - \frac{1}{r})^2 = a^2$,so $r - \frac{1}{r} = a$ (taking the positive root for distance).$r^2 - ar - 1 = 0$.Solving for $r$ using the quadratic formula,$r = \frac{a + \sqrt{a^2 + 4}}{2}$.

The maximum distance from the origin of coordinates to the point $z$ satisfying the equation $\left| z + \frac{1}{z} \right| = a$ is

Similar Questions

Let $z = x + iy$,where $x$ and $y$ are real. The points $(x, y)$ in the $X-Y$ plane for which $\frac{z+i}{z-i}$ is purely imaginary,lie on

Let $z$ be the complex number satisfying $|z-5| \le 3$ and having the maximum positive principal argument. Then $34|\frac{5z-12}{5iz+16}|^{2}$ is equal to:

Let $z$ be a complex number such that the real part of $\frac{z-2i}{z+2i}$ is zero. Then,the maximum value of $|z-(6+8i)|$ is equal to:

Let complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $(x-x_0)^2+(y-y_0)^2=r^2$ and $(x-x_0)^2+(y-y_0)^2=4r^2$,respectively. If $z_0=x_0+iy_0$ satisfies the equation $2|z_0|^2=r^2+2$,then $|\alpha|=$

The equation of the locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$,where $z=x+iy$ is a complex number,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module