Let z be the complex number satisfying |z − 5| ≤ 3 and having maximum positive principal argument. Then 34|(5z−12)/(5iz+16)|² is equal t

Every complex number $z = x + iy$ corresponds to the point $(x,y)$ in the Argand plane (complex plane). This geometric representation is fundamental to solving complex number problems in JEE.

Modulus: $|z| = \sqrt{x^2+y^2}$ = distance from origin to point $(x,y)$

Argument: $\arg(z) = \tan^{-1}(y/x)$ = angle made with positive x-axis

Principal Argument: The argument in the range $(-\pi, \pi]$. Positive principal argument means the point is in the upper half plane ($y > 0$) or on positive x-axis.

$|z – z_0| = r$ represents a circle centred at $z_0$ with radius $r$.
$|z – z_0| \leq r$ represents the closed disk (circle + interior).

Common forms to recognise:
• $|z – a| = |z – b|$: perpendicular bisector of $a$ and $b$
• $|z – a| + |z – b| = k$: ellipse with foci $a,b$
• $|z – a| / |z – b| = k$: circle (Apollonius circle) if $k \neq 1$
• $\arg(z – a) = \theta$: ray from $a$ at angle $\theta$

To find the complex number $z$ on/inside a circle with maximum argument:

Step 1: Draw tangent lines from the origin to the circle boundary.
Step 2: The tangent point in the upper half plane gives maximum argument.
Step 3: Use right triangle: if centre is at distance $d$ from origin and radius is $r$, then: $$\sin(\theta_{max}) = \frac{r}{d}, \quad \text{tangent length} = \sqrt{d^2 – r^2}$$
Here: $d = |OC| = 5$, $r = 3$, so tangent length $= \sqrt{25-9} = 4$.
The tangent point: $z = 4e^{i\theta} = 4(\cos\theta + i\sin\theta)$ where $\sin\theta = 3/5$.

A key property of complex modulus: $$\left|\frac{z_1}{z_2}\right|^2 = \frac{|z_1|^2}{|z_2|^2}$$ This allows us to compute the modulus squared of a quotient by computing numerator and denominator moduli separately.

Also remember:
• $|z_1 z_2| = |z_1| \cdot |z_2|$
• $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2\text{Re}(z_1\overline{z_2})$
• $|z|^2 = z\bar{z}$
• $|iz| = |i| \cdot |z| = |z|$ (multiplying by $i$ only rotates, doesn’t change modulus)

Multiplying a complex number $z$ by $i$ rotates it by $90°$ anticlockwise: $$i \cdot z = i(x+iy) = ix + i^2y = -y + ix$$ So the point $(x,y)$ maps to $(-y, x)$.

In this problem: $5iz = i(16+12i) = 16i – 12 = -12 + 16i$. This is just a $90°$ rotation of $16+12i$.

Powers of $i$:
$i^0=1,\ i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$ (period 4)

We must verify $|z-5| = 3$: $$z – 5 = \frac{16+12i}{5} – 5 = \frac{16+12i-25}{5} = \frac{-9+12i}{5}$$ $$|z-5| = \frac{\sqrt{81+144}}{5} = \frac{\sqrt{225}}{5} = \frac{15}{5} = 3 \ ✓$$ The point lies exactly on the boundary circle, confirming it is the tangent point.

❌ Mistake 1: Confusing $|z-5|\leq 3$ (disk) with $|z-5|=3$ (circle). The maximum argument is on the boundary circle, not the interior.

❌ Mistake 2: Thinking maximum argument occurs at the topmost point of the circle $(5, 3)$. This is not true — the maximum argument occurs at the tangent point from the origin, not the topmost point.

❌ Mistake 3: Computing $5iz$ as $5i\cdot z$ incorrectly. Remember $i^2 = -1$, so $i(12i) = 12i^2 = -12$ (real, not imaginary).

❌ Mistake 4: Forgetting that tangent length $= \sqrt{d^2-r^2}$, not $d-r$.

❌ Mistake 5: Computing $|5z-12|^2$ as $|5z|^2 – |12|^2$. Modulus does NOT distribute over subtraction like that. Always compute $|a+bi|^2 = a^2+b^2$ directly.

$|z-z_0|\leq r$: disk, centre $z_0$ Max arg: tangent from origin $\sin\theta=r/d$, tangent$=\sqrt{d^2-r^2}$ $z_{tangent}=\sqrt{d^2-r^2}\cdot e^{i\theta}$ $|a+bi|^2=a^2+b^2$ $|z_1/z_2|^2=|z_1|^2/|z_2|^2$ $i^2=-1$, $i(a+bi)=-b+ai$

Complex number geometry is a high-weightage topic in JEE Main — typically 2–3 questions per session. Problems involving maximum/minimum argument, locus on Argand plane, and modulus expressions appear regularly. The key skill tested here is:

(1) Converting $|z-5|\leq3$ to a circle
(2) Finding the tangent point for maximum argument
(3) Clean computation of complex modulus squared

This 3-step approach works for all “maximum/minimum argument” problems. Time target: 3 minutes in JEE Main.