Super Duper Magnitudes

Let ${ Z }_{ 1 }=\quad 3i\quad \quad \& \quad \quad { Z }_{ 2 }\quad =\quad 4$ .

By Triangle Inequality :

$\left| Z-{ Z }_{ 1 } \right| +\quad \left| Z-{ Z }_{ 2 } \right| \quad \ge \quad \left| { Z }_{ 1 }{ +Z }_{ 2 } \right|$ .

But Note Here it is : $\left| Z\quad -\quad { Z }_{ 1 } \right| +\quad \left| Z\quad -{ \quad Z }_{ 2 } \right| \quad =\quad \left| { Z }_{ 1 }{ \quad +\quad Z }_{ 2 } \right|$ .

So

${ Z }_{ 1 }\quad ,\quad { Z }_{ 2 }\quad ,\quad Z\quad \rightarrow \quad Collinear$ .

So " Z " Lies on line L :

$3x\quad +\quad 4y\quad =\quad 12$ .

Now ${ \left| Z \right| }_{ min }$ . is eaual to perpendicular drop from origin To the Line "L"

${ \left| Z \right| }_{ min }\quad =\quad \cfrac { 12 }{ 5 }$ .

Let $z=a+bi$ and substitute to the equation then we have

$|a+(b-3)i|+|(a-4)+bi|=5$

The absolute value of a complex number $a+bi$ is $\sqrt{a^2+b^2}$ . So, we can rewrite the equation to

$\sqrt{a^2+(b-3)^2}+\sqrt{(a-4)^2+b^2}=5$

and our goal is to solve for the minimum value of $\sqrt{a^2+b^2}$ .

Now, take a look at this picture:

triangle

$AP: \sqrt{a^2+(b-3)^2}$

$PB: \sqrt{(a-4)^2+b^2}$

Since $AP+PB=AB=5$ , it yields that point $P$ lies on line $AB$ .

Noticed that $OP=\sqrt{a^2+b^2}$ is the expression which minimum value we seek. In other words, we must find the shortest length of line $OP$ .

Now the problem gets elementary, the shortest possible length of line $OP$ is when it is perpendicular to $AB$ . By the area of the triangle, we have

$\frac{1}{2}\times 3 \times 4=\frac{1}{2}\times 5 \times OP$

$OP=2.4$ .