JEE Mathematics 2023 Question with Solution

Given: Two non-zero complex numbers $z_1$ and $z_2$ satisfy

$$\operatorname{Re}(z_1 z_2)=0$$

and

$$\operatorname{Re}(z_1+z_2)=0$$

Find: Which sign combinations of $\operatorname{Im}(z_1)$ and $\operatorname{Im}(z_2)$ are possible.

Let

$$z_1=x_1+i y_1, \qquad z_2=x_2+i y_2$$

where $x_1, x_2$ are real parts and $y_1, y_2$ are imaginary parts.

Now,

$$z_1 z_2=(x_1+i y_1)(x_2+i y_2)$$

So,

$$\operatorname{Re}(z_1 z_2)=x_1 x_2-y_1 y_2=0$$

which gives

$$x_1 x_2=y_1 y_2$$

Also,

$$\operatorname{Re}(z_1+z_2)=x_1+x_2=0$$

so

$$x_1=-x_2$$

Hence $x_1$ and $x_2$ are opposite in sign, therefore their product $x_1 x_2$ is negative when both are non-zero. Since

$$y_1 y_2=x_1 x_2$$

we get that $y_1 y_2$ must also be negative. Therefore $y_1$ and $y_2$ have opposite signs.

So the possible cases are:

• $\operatorname{Im}(z_1)<0$ and $\operatorname{Im}(z_2)>0$
• $\operatorname{Im}(z_1)>0$ and $\operatorname{Im}(z_2)<0$

Therefore, the correct options are B and C. The solution working supports option-set (2), although the solution incorrectly states option A.

Given:

$$\operatorname{Re}(z_1 z_2)=0, \qquad \operatorname{Re}(z_1+z_2)=0$$

Find: The valid sign pattern of the imaginary parts.

From

$$x_1+x_2=0$$

we write

$$x_2=-x_1$$

Substitute into

$$x_1 x_2=y_1 y_2$$

Then

$$x_1(-x_1)=y_1 y_2$$

which gives

$$-x_1^2=y_1 y_2$$

If $x_1\neq 0$, then

$$y_1 y_2<0$$

so $y_1$ and $y_2$ must be of opposite signs.

Thus both imaginary parts cannot be positive together, and both cannot be negative together. Only the mixed-sign cases remain valid.

Therefore, the correct answer is option-set (2): B and C.