Let z be a complex number such that |z+2| = |z-2| and arg((z+3)/(z-i)) = pi/4. Then |z|^2 is equal to:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q3 – JEE-MAIN 2026

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Home Maths Complex & QuadraticQuestion #72

Let

z

be a complex number such that

| z + 2 | = | z - 2 |

and

\arg (\frac{z + 3}{z - i}) = \frac{π}{4}

. Then

| z |^{2}

is equal to:

Clue 1 of 3: Geometric Interpretation

The condition $| z - a | = | z - b |$ implies that $z$ lies on the perpendicular bisector of the line segment joining $a$ and $b$ .

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine the locus of

z

from the first condition✦ Active

The condition $| z + 2 | = | z - 2 |$ means that the distance of $z$ from $- 2$ is equal to its distance from $2$ . Geometrically, this implies that $z$ lies on the perpendicular bisector of the segment joining $- 2$ and $2$ on the real axis. This perpendicular bisector is the imaginary axis. Therefore, $z$ must be a purely imaginary number, so we can write $z = i y$ for some real number $y$ .

Step 2: Substitute

z = i y

into the argument condition and simplify○ Expand

Substitute $z = i y$ into the second condition:

\arg (\frac{i y + 3}{i y - i}) = \frac{π}{4}

Factor out $i$ from the denominator and simplify the fraction:

\arg (\frac{3 + i y}{i (y - 1)}) = \frac{π}{4}

To simplify the complex fraction, multiply the numerator and denominator by $- i$ :

\arg (\frac{(3 + i y) (- i)}{i (y - 1) (- i)}) = \arg (\frac{- 3 i - i^{2} y}{- (y - 1) i^{2}}) = \arg (\frac{y - 3 i}{y - 1}) = \frac{π}{4}

Let $w = \frac{y - 3 i}{y - 1}$ . We can write $w$ in the form $X + i Y$ :

w = \frac{y}{y - 1} - \frac{3}{y - 1} i

💡 Teacher's Secret Hint

Remember that $i^{2} = - 1$ . Be careful with signs when multiplying by $- i$ .

Step 3: Solve for

y

and calculate

| z |^{2}

○ Expand

Given $\arg (w) = \frac{π}{4}$ , this implies that the real part of $w$ must be equal to its imaginary part, and both must be positive (since $\frac{π}{4}$ is in the first quadrant). So, $X = Y > 0$ .

\frac{y}{y - 1} = - \frac{3}{y - 1}

Since $y - 1 \neq 0$ (otherwise the expression is undefined), we can multiply both sides by $y - 1$ :

y = - 3

Now, we check if $X > 0$ and $Y > 0$ for $y = - 3$ :

X = \frac{- 3}{- 3 - 1} = \frac{- 3}{- 4} = \frac{3}{4}

Y = - \frac{3}{- 3 - 1} = - \frac{3}{- 4} = \frac{3}{4}

Both $X$ and $Y$ are positive and equal, so $y = - 3$ is a valid solution. Therefore, $z = - 3 i$ .

Finally, we need to find $| z |^{2}$ :

| z |^{2} = | - 3 i |^{2} = (0)^{2} + (- 3)^{2} = 9

💡 Teacher's Secret Hint

Always verify the quadrant condition for the argument. For $\arg (X + i Y) = \frac{π}{4}$ , both $X$ and $Y$ must be positive.

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Maths Question 3 – JEE-MAIN 2026

Clue 1 of 3: Geometric Interpretation

🧠 Full Solution Path