Animated Solution for Mathematics - Complex Numbers: Let $u=\frac{2z+i}{z-ki}$, $z=x+iy$ and $k>0$. If the curve represented by $\text{Re}(u)+\text{Im}(u)=1$ intersects the y-axis at the point $P$ and $Q$ where $PQ=5$, then the value of $K$ is :

Imagine you are standing on the complex plane, a vast, two-dimensional canvas where every point is a complex number $z=x+iy$. We are given a transformation:

where $k>0$. This is not just an algebraic expression; it is a mapping that transforms the geometry of the plane. Our mission is to uncover the path defined by the condition $\text{Re}(u)+\text{Im}(u)=1$.

Rationalization

To understand this curve, we must first break $u$ into its real and imaginary parts. We substitute $z=x+iy$ into the expression for $u$:

We cannot easily separate the real and imaginary parts with $i$ in the denominator. We perform a surgical procedure: rationalization. We multiply the numerator and denominator by the conjugate of the denominator, $x-i(y-k)$.

This yields:

Now, the denominator is purely real. We can extract the real and imaginary parts as follows:

The problem gives us the condition $\text{Re}(u)+\text{Im}(u)=1$. Substituting our expressions, we have:

Multiplying both sides by the denominator and expanding, we get:

Rearranging these terms, we arrive at the elegant equation:

This is the equation of a circle. We have successfully translated a complex transformation into a familiar geometric shape.

The problem states that this circle intersects the y-axis at points $P$ and $Q$. On the y-axis, the x-coordinate is always zero. Substituting $x=0$ into our circle equation, we get:

This is a quadratic equation in $y$. Let its roots be $y_1$ and $y_2$, which represent the y-coordinates of $P$ and $Q$. We are given that the distance $PQ=5$, which means $|y_1-y_2|=5$.

Using the identity $(y_1-y_2{)}^2=(y_1+y_2{)}^2-4y_1{y}_2$ and applying Vieta's relations where $y_1+y_2=-1$ and $y_1{y}_2=-(k^2+k)$, we get:

This simplifies to:

Dividing by 4, we obtain $k^2+k-6=0$. Factoring this, we find $(k+3)(k-2)=0$, giving $k=-3$ or $k=2$.

Since the problem specifies $k>0$, we reject $-3$. We conclude that the final answer is $k=2$.