Animated Solution for Mathematics - Complex Numbers: Let $A=\{z\in \mathbb{C}:|z-2-i|=3\}$, $B=\{z\in \mathbb{C}:\text{Re}(z-iz)=2\}$ and $S=A\cap B$. Then ${\sum }_{z\in S}|z{|}^2$ is equal to ________ .

Complex numbers are often misunderstood as abstract entities, but they are deeply geometric. We are given two sets, $A$ and $B$, on the Argand plane.

Set $A$ is defined by $|z-(2+i)|=3$. This represents a circle with center at $2+i$, corresponding to the point $(2,1)$ in the Cartesian plane, and a radius of $3$.

Set $B$ is defined by $\text{Re}(z-iz)=2$. By substituting $z=x+iy$, we find that $z-iz=(x+y)+i(y-x)$. The real part is simply $x+y$, so Set $B$ is the line $x+y=2$.

We need to find the intersection set $S=A\cap B$. The circle is given by the equation:

Substituting the line equation $y=2-x$ into the circle equation, we obtain:

Simplifying this expression leads to:

Expanding the squares, we arrive at:

Combining like terms, we get $2x^2-6x+5=9$, which simplifies to $2x^2-6x-4=0$. Dividing by $2$, we obtain the quadratic equation:

Let the roots of the quadratic be $x_1$ and $x_2$. From Vieta's relations, we know:

We need to find the sum of the squared magnitudes $\sum |z{|}^2$. Since $|z{|}^2=x^2+y^2$ and $y=2-x$, we have:

The sum we seek is $(2x_1^2-4x_1+4)+(2x_2^2-4x_2+4)$, which simplifies to:

We calculate the sum of squares using the identity $x_1^2+x_2^2=(x_1+x_2{)}^2-2x_1{x}_2$. Substituting our Vieta values:

Now, plug this back into our sum expression:

The final result is 22.