Animated Solution for Mathematics - Complex Numbers: Let $z_1=10+6i$ and $z_2=4+6i$. If $z$ is any complex number such that the argument of $\frac{z-z_1}{z-z_2}$ is $\frac{\pi }{4}$, then prove that $|z-7-9i|=3\sqrt{2}$.

Unveiling the Locus

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are exploring the hidden geometry of the complex plane.

When you see an expression like $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{4}$, do not let the algebra intimidate you. Instead, close your eyes and visualize the Argand plane. This is not just a collection of numbers; it is a canvas where complex numbers dance in perfect geometric harmony.

The Chord of Connection

Let us begin by plotting our points. We are given $z_1=10+6i$ and $z_2=4+6i$.

In the Argand plane, these are simply the coordinates $A(10,6)$ and $B(4,6)$. Look closely—do you see it? Both points share the same imaginary part, $6$.

This means they sit perfectly on a horizontal line. Imagine drawing a chord connecting these two points. This chord is the foundation of our entire problem, acting as the fixed segment that defines the arc we are about to discover.

The Locus Mystery

Now, consider the condition $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{4}$. In the language of complex numbers, the argument of a quotient is the difference of the arguments:

This represents the angle subtended by the segment $z_1{z}_2$ at the point $z$.

From the ancient wisdom of circle geometry, we know that the locus of points that subtend a constant angle to a fixed chord is an arc of a circle. Because our angle is $\frac{\pi }{4}$ (which is positive and acute), the point $z$ traces out the major arc of a circle passing through $z_1$ and $z_2$.

The Geometry of the Center

To define this circle, we need its center $C$ and its radius $R$. We know that the center must lie on the perpendicular bisector of our chord $z_1{z}_2$.

Since our chord is horizontal, the perpendicular bisector is the vertical line passing through the midpoint of $z_1$ and $z_2$. The midpoint $M$ is:

Thus, the center $C$ must have coordinates $(7,k)$. Now, let us invoke the Inscribed Angle Theorem. If the angle subtended by the chord at the circumference is $\frac{\pi }{4}$, then the angle subtended at the center must be double that: $\frac{\pi }{2}$.

This is a right angle! If we draw the triangle $△CM{z}_1$, where $M$ is the midpoint, the perpendicular bisector splits this central angle in half. Therefore, $\angle MC{z}_1=\frac{\pi }{4}$.

The Final Proof

We are now in the home stretch. In our right-angled triangle $△CM{z}_1$, we have:

The distance $M{z}_1$ is simply the horizontal distance from the midpoint $(7,6)$ to $(10,6)$, which is $10-7=3$. Since $\tan (\frac{\pi }{4})=1$, we have:

Since the center is at $(7,k)$ and the midpoint is at $(7,6)$, the vertical distance $CM$ is $|k-6|=3$. This gives us $k=9$. Our center is $C(7,9)$.

Finally, the radius $R$ is the hypotenuse of $△CM{z}_1$:

The equation of a circle in the complex plane is $|z-C|=R$. Substituting our center $C=7+9i$ and radius $R=3\sqrt{2}$, we arrive at:

Or, written as requested:

Look at that! The geometry and the algebra have converged perfectly. You have successfully navigated the locus, understood the circle, and arrived at the proof. Remember, in complex numbers, geometry is your greatest ally.