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

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