Let z ∈ C be such that (z² + 3i)/(z - 2 + i) = 2 + 3i. Then the sum of all possible values of z² is

Q: Wait, why can't we just solve for $z$ using the quadratic formula?

You absolutely could! However, in JEE Advanced, time is your most precious resource. Solving for $z$ would involve finding the roots of a complex quadratic, which often leads to messy square roots of complex numbers. Using Vieta's formulas allows you to jump straight to the sum of squares without ever needing to know the individual values of $z_1$ and $z_2$.

Q: Is it always safe to assume there are two roots?

Yes, for a quadratic equation of the form $az^2 + bz + c = 0$ where $a \neq 0$, the Fundamental Theorem of Algebra guarantees exactly two roots in the complex plane (counting multiplicity). Even if the roots are identical, the identity $z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1z_2$ still holds perfectly.

Q: Why does $i^2$ become $-1$ in the middle of the calculation?

That is the fundamental definition of the imaginary unit $i$. Whenever you see $i^2$, it is a signal to replace it with $-1$. This is the most common place where students lose marks due to a sign error. Always pause, check your $i^2$ terms, and convert them to real numbers before proceeding.

Question

Let $z \in \mathbb{C}$ be such that $\frac{z^2 + 3i}{z - 2 + i} = 2 + 3i$. Then the sum of all possible values of $z^2$ is

Options:

(A) $19 - 2i$

(B) $-19 - 2i$

(C) $19 + 2i$

(D) $-19 + 2i$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex equation:

$$\frac{z^2 + 3i}{z - 2 + i} = 2 + 3i$$

At first glance, it looks like a daunting rational expression. In the realm of JEE Advanced, complexity is often just a veil for simplicity. Our goal is to find the sum of all possible values of $z^2$.

### The Art of Cross-Multiplication

The denominator is the source of our friction. To liberate the equation, we perform a simple, yet powerful, act of cross-multiplication.

By bringing the term $(z - 2 + i)$ to the right-hand side, we transform the rational expression into a linear-looking product:

$$z^2 + 3i = (z - 2 + i)(2 + 3i)$$

Now, the equation is laid bare, ready for expansion.

### The Expansion and the $i^2$ Trap

Expanding the right-hand side requires precision. We distribute $z$ and the constant terms carefully:

$$z^2 + 3i = z(2 + 3i) - 2(2 + 3i) + i(2 + 3i)$$

As we calculate these terms, we must be vigilant. When we encounter $i(3i)$, we get $3i^2$. Remember, the soul of complex algebra lies in the identity $i^2 = -1$.

Thus, $3i^2$ becomes $-3$. By simplifying, we arrive at:

$$z^2 + 3i = (2 + 3i)z - 4 - 6i + 2i - 3$$

Combining the real and imaginary parts, we get:

$$z^2 + 3i = (2 + 3i)z - 7 - 4i$$

### The Quadratic Transformation

Now, we shift everything to the left-hand side to reveal the standard quadratic form $az^2 + bz + c = 0$.

Moving the terms, we get:

$$z^2 - (2 + 3i)z + 3i + 7 + 4i = 0$$

This simplifies beautifully to:

$$z^2 - (2 + 3i)z + 7 + 7i = 0$$

This is the heart of the problem. We have a quadratic equation where the coefficients are complex numbers.

### The Elegance of Vieta's Formulas

Why solve for $z$ when we can solve for the sum of squares directly? Let the roots be $z_1$ and $z_2$.

Vieta's formulas tell us that the sum of the roots is:

$$z_1 + z_2 = -\frac{b}{a} = 2 + 3i$$

The product of the roots is:

$$z_1z_2 = \frac{c}{a} = 7 + 7i$$

We need the sum of the squares, $z_1^2 + z_2^2$. We use the classic identity:

$$z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1z_2$$

### The Final Calculation

Now, we substitute our known values:

$$z_1^2 + z_2^2 = (2 + 3i)^2 - 2(7 + 7i)$$

First, expand $(2 + 3i)^2$:

$$2^2 + (3i)^2 + 2(2)(3i) = 4 - 9 + 12i = -5 + 12i$$

Next, calculate $-2(7 + 7i) = -14 - 14i$. Finally, combine them:

$$(-5 + 12i) + (-14 - 14i) = (-5 - 14) + (12i - 14i) = -19 - 2i$$

The final result is **$-19 - 2i$**. You have successfully navigated the complexity and arrived at the truth.

Animated Solution for Mathematics - Complex Numbers: Let $z \in C$ be such that $\frac{z ^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is

Select Answer:

Visualized Solution

Analyze the Given Equation

Eliminate the Denominator

Expand the Right-Hand Side

Perform Atomic Multiplication

Simplify the Right-Hand Side

Form the Quadratic Equation $a z^{2} + b z + c = 0$

Identify Roots and Vieta's Formulas

The Sum of Squares Formula $z_{1}^{2} + z_{2}^{2}$

Substitute the Values

Square the Complex Sum $(2 + 3 i)^{2}$

Multiply the Product Term

Final Calculation

Conclusion and Key Takeaways

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Art of Cross-Multiplication

The Expansion and the $i^{2}$ Trap

The Quadratic Transformation

The Elegance of Vieta's Formulas

The Final Calculation

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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−2+iz2+3i​=2+3i. Then the sum of all possible values of z2 is