Animated Solution for Mathematics - Complex Numbers: Let $z$ be a complex number such that $|z-6|=5$ and $|z+2-6i|=5$. Then the value of $z^3+3z^2-15z+141$ is equal to

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant architecture of the complex plane. Many students look at a problem involving complex numbers and immediately reach for the substitution $z=x+iy$.

While that is a valid path, the true masters of JEE Advanced look for the geometry hidden beneath the algebra. Let us embark on this journey together.

We begin with two conditions: $|z-6|=5$ and $|z+2-6i|=5$. In the Argand plane, the expression $|z-z_0|=r$ is the definition of a circle with center $z_0$ and radius $r$.

Our first condition, $|z-6|=5$, tells us we have a circle centered at $(6,0)$ with a radius of $5$. Our second condition, $|z-(-2+6i)|=5$, reveals a second circle centered at $(-2,6)$ with the same radius of $5$.

Imagine these two circles on your graph paper. They are dancing around each other. The complex number $z$ must satisfy both, meaning it must lie at the intersection of these two circles.

Now, we must find where they meet. We set $z=x+iy$. The first circle becomes $(x-6{)}^2+y^2=25$, which expands to $x^2-12x+36+y^2=25$, or:

The second circle becomes $(x+2{)}^2+(y-6{)}^2=25$, which expands to $x^2+4x+4+y^2-12y+36=25$, or:

Here is the beauty of the system: if we subtract the first equation from the second, the quadratic terms $x^2$ and $y^2$ vanish entirely. We are left with $16x-12y+4=0$, which simplifies to $4x-3y+1=0$. This is the common chord of the two circles.

By expressing $x$ as $x=\frac{3y-1}{4}$ and substituting it back into our first circle equation, we find that the resulting quadratic in $y$ is $(y-3{)}^2=0$. This implies $y=3$, and consequently, $x=2$. Our intersection point is $z=2+3i$.

Now, we face the final challenge: evaluating $P(z)=z^3+3z^2-15z+141$. A novice would plug $z=2+3i$ into this cubic expression. That is a path to disaster, filled with complex arithmetic errors.

Instead, we use the 'JEE Secret Weapon.' We know $z=2+3i$, which means $z-2=3i$. Squaring both sides gives $(z-2{)}^2=(3i{)}^2$, which simplifies to $z^2-4z+4=-9$, or:

This quadratic is the 'DNA' of our complex number. Any polynomial involving $z$ can be divided by this quadratic. When we perform polynomial long division of $P(z)$ by $(z^2-4z+13)$, we get a quotient of $(z+7)$ and a remainder of $50$.

Thus, we can write the expression as:

Since we know $z^2-4z+13=0$, the entire first term vanishes into thin air. We are left with the remainder: 50. This is the elegance of mathematics. We did not need to calculate the cube of a complex number; we simply understood the structure of the expression.