Let z ∈ C be such that |z|

Q: Why can't we just substitute $z = x + iy$ into the original equation and solve it that way?

You certainly could, but it would lead to a nightmare of algebraic expansion and complex conjugate manipulation. The inverse mapping technique is the elegant, professional way to handle Möbius transformations. It allows us to transform the problem into a simple geometric inequality, which is much faster and less prone to calculation errors.

Q: How do I know the boundary is a line and not a circle?

In a Möbius transformation, the boundary of the unit disk $|z|=1$ maps to a line if the denominator vanishes on the boundary. Here, the denominator $5(1-z)$ vanishes at $z=1$, which is on the boundary $|z|=1$. Thus, the boundary maps to a straight line, specifically the perpendicular bisector of the points $1$ and $-\frac{3}{5}$.

Q: Why is the final answer $5\text{Re}(\omega) > 1$ and not $5\text{Re}(\omega) < 1$?

Great catch! To determine the correct side of the boundary, test a point inside the original disk, such as $z=0$. If $z=0$, then $\omega = \frac{5+3(0)}{5(1-0)} = 1$. Since $5\text{Re}(1) = 5$, and $5 > 1$, the inequality must be greater than $1$.

Question

Let $z \in C$ be such that $|z| < 1$. If $\omega = \frac{5+3z}{5(1-z)}$, then :- Options: (A) $5 ext{Im}(\omega) < 1$ (B) $4 ext{Im}(\omega) > 5$ (C) $5 ext{Re}(\omega) > 1$ (D) $5 ext{Re}(\omega) > 4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup We are given a complex number $z$ such that $|z| < 1$ and a Möbius transformation defined by: $$\omega = \frac{5+3z}{5(1-z)}$$ Our goal is to determine the region in the $\omega$-plane that corresponds to the interior of the unit disk in the $z$-plane. ### The Art of Inverse Mapping The most common mistake is attempting to substitute $z$ directly. Instead, we use the power of inverse mapping to view the transformation from the perspective of $\omega$. Starting with the transformation, we cross-multiply: $$5\omega(1-z) = 5+3z$$ Expanding this, we obtain: $$5\omega - 5\omega z = 5+3z$$ Next, we group the terms involving $z$ on one side: $$5\omega - 5 = 3z + 5\omega z$$ Factoring out $z$, we find: $$5\omega - 5 = z(3 + 5\omega)$$ Finally, we isolate $z$ to express it as a function of $\omega$: $$z = \frac{5\omega - 5}{5\omega + 3}$$ ### The Geometric Leap We apply the given condition $|z| < 1$ to our inverse expression: $$\left| \frac{5\omega - 5}{5\omega + 3} ight| < 1$$ Using the property that the modulus of a quotient is the quotient of the moduli, we have: $$\frac{|5\omega - 5|}{|5\omega + 3|} < 1$$ Since the modulus is always positive, we multiply the denominator across: $$|5\omega - 5| < |5\omega + 3|$$ Dividing both sides by $5$ yields the simplified inequality: $$|\omega - 1| < |\omega + \frac{3}{5}|$$ ### The Final Revelation This inequality represents the set of points $\omega$ that are closer to $1$ than to $-\frac{3}{5}$ in the complex plane. The boundary of this region is the perpendicular bisector of the segment connecting $1$ and $-\frac{3}{5}$. The midpoint of this segment is: $$\frac{1 - \frac{3}{5}}{2} = \frac{\frac{2}{5}}{2} = \frac{1}{5}$$ The boundary is therefore the vertical line $ ext{Re}(\omega) = \frac{1}{5}$. Since $\omega$ is closer to $1$, it must lie to the right of this line. **Conclusion:** The region is defined by **$ ext{Re}(\omega) > \frac{1}{5}$**, which can also be written as **$5 ext{Re}(\omega) > 1$**.

Animated Solution for Mathematics - Complex Numbers: Let $z \in C$ be such that $∣ z ∣ < 1$ . If $ω = \frac{5 + 3 z}{5 ( 1 - z )}$ , then :-

Select Answer:

Visualized Solution

The Given Condition

Strategy: Inverse Mapping

Cross-Multiplying

Expanding the Terms

Grouping $z$ Terms

Factoring $z$

Isolating $z$

Applying the Modulus Condition

Simplifying the Modulus

Standardizing the Inequality

Geometric Interpretation

The Perpendicular Bisector

Identifying the Region

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Art of Inverse Mapping

The Geometric Leap

The Final Revelation

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z∈C be such that ∣z∣<1. If ω=5(1−z)5+3z​, then :-