Let z in mathbb{C} be such that |z|

Question

(C) Given $w = \frac{5 + 3z}{5(1 - z)}$.Rearranging to solve for $z$:$5w(1 - z) = 5 + 3z$$5w - 5wz = 5 + 3z$$5w - 5 = z(3 + 5w)$$z = \frac{5w - 5}{5w

Vedclass · Accepted Answer

$5 	ext{ Re}(w) > 1$

Vedclass · Answer

(C) Given $w = \frac{5 + 3z}{5(1 - z)}$.Rearranging to solve for $z$:$5w(1 - z) = 5 + 3z$$5w - 5wz = 5 + 3z$$5w - 5 = z(3 + 5w)$$z = \frac{5w - 5}{5w + 3}$.Since $|z| $|5w - 5| Divide by $5$:$|w - 1| This represents the set of points $w$ that are closer to $1$ than to $-\frac{3}{5}$ in the complex plane.The perpendicular bisector of the segment joining $1$ and $-\frac{3}{5}$ is the line $x = \frac{1 - 3/5}{2} = \frac{2/5}{2} = \frac{1}{5}$.Since the points must be closer to $1$,we have $	ext{Re}(w) > \frac{1}{5}$,which implies $5 	ext{ Re}(w) > 1$.

Let $z \in \mathbb{C}$ be such that $|z| < 1$. If $w = \frac{5 + 3z}{5(1 - z)}$,then

Similar Questions

For any complex number $z$,the minimum value of $|z|+|z-1|$ is

If $|z_1| = |z_2|$ and $\arg\left( \frac{z_1}{z_2} \right) = \pi$,then $z_1 + z_2$ is equal to

If $z$ is a complex number such that $|z+4| \geq 3$,then the smallest value of $|z+3|$ is

Let ${z_1}, {z_2}, {z_3}$ be three vertices of an equilateral triangle circumscribing the circle $|z| = \frac{1}{2}$. If ${z_1} = \frac{1}{2} + \frac{\sqrt{3}i}{2}$ and ${z_1}, {z_2}, {z_3}$ are in anticlockwise sense,then ${z_2}$ is

If $z$ is a complex number satisfying $|z - 3| \leq 5$,then the range of $|z + 3i|$ is (where $i = \sqrt{-1}$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module