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

If $z$ is a complex number such that $\frac{z-i}{z-1}$ is purely imaginary,then the minimum value of $|z-(3+3i)|$ is:

If the least and the largest real values of $\alpha,$ for which the equation $z+\alpha|z-1|+2i=0$ ($z \in \mathbb{C}$ and $i=\sqrt{-1}$) has a solution,are $p$ and $q$ respectively; then $4(p^2+q^2)$ is equal to ..........

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

Let $S = \{z \in \mathbb{C} : \left|\frac{z-6i}{z-2i}\right| = 1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right| = \frac{3}{5}\}$. Then $\sum_{z \in S} |z|^2$ is equal to

If $z=x+iy$ represents a point $P$ in the Argand plane, then the area of the region represented by the inequality $2 < |z-(1+i)| < 3$ is (in $\pi$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module