JEE Mathematics 2025 Question with Solution

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

Find: The sum of all possible values of $z$.

From the solution, the working concludes with the quadratic

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

Let its roots be $z_1$ and $z_2$. Then

$$z_1 + z_2 = 2 + 3i, \quad z_1 z_2 = 7 + 7i$$

Now,

$$\begin{aligned} z_1^2 + z_2^2 &= (z_1 + z_2)^2 - 2z_1 z_2 \\ &= (2 + 3i)^2 - 2(7 + 7i) \\ &= 4 + 12i - 9 - 14 - 14i \\ &= -19 - 2i \end{aligned}$$

Therefore, the sum of all possible values of $z$ is $-19 - 2i$, so the correct option is B.

The solution is internally inconsistent because the original linear equation gives a unique value of $z$, but the provided the solution explicitly concludes the asked sum as $-19 - 2i$. By the answer-resolution rule, the solution is treated as authoritative.

Relevant extracted algebra shown in the solution:

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

which was expanded there as

$$z + 3i = 2z + 3iz - 7 - 4i$$

and rearranged to

$$z(1 + 3i) = 7 + 7i$$

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

Find: The sum of all possible values of $z$.

The solution first clears the denominator:

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

Then it expands:

$$\begin{aligned} z + 3i &= 2(z - 2 + i) + 3i(z - 2 + i) \\ &= 2z - 4 + 2i + 3iz - 6i + 3i^2 \\ &= 2z + 3iz - 7 - 4i \end{aligned}$$

So,

$$z + 3i - 2z - 3iz + 7 + 4i = 0$$

which gives

$$-z - 3iz + 7 + 7i = 0$$

and hence

$$z(1 + 3i) = 7 + 7i$$

After this, the solution rewrites the situation as the quadratic

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

Let the roots be $z_1$ and $z_2$. Then

$$z_1 + z_2 = 2 + 3i, \quad z_1 z_2 = 7 + 7i$$

Using the identity from the provided solution,

$$\begin{aligned} z_1^2 + z_2^2 &= (z_1 + z_2)^2 - 2z_1 z_2 \\ &= (2 + 3i)^2 - 2(7 + 7i) \\ &= -19 - 2i \end{aligned}$$

Therefore, according to the solution, the required answer is $-19 - 2i$, which corresponds to Option B.