Algebra with complex numbers Let $m$ be the minimum value of $|z|$, where $z$ is a complex number such that
$$ |z-3i | + |z-4 | = 5. $$
Then $m$ can be written in the form $\dfrac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find the value of $a+b$.
I cannot seem to get rid of the absolute values, making it impossible to solve this.  Is it possible for someone to give me some help?
Thank you!
 A: Given $|z-4|+|z-3i|=5\;,$ Now Put $z=x+iy\;,$ We get $$\displaystyle \sqrt{(x-4)^2+y^2}+\sqrt{x^2+(y-3)^2}=5$$
Now Using Minkowski inequality, We get
$$\sqrt{(4-x)^2+y^2}+\sqrt{x^2+(3-y)^2}\geq \sqrt{[(4-x)-x]^2+[y-(3-y)]^2}=5$$
And equality Hold when $$\frac{4-x}{x}=\frac{y}{3-y}\Rightarrow 12-3x-4y+xy=xy$$
So we get $3x+4y=12$
So here we have equality condition is hold.
Now Using Cauchy Schwartz Inequality
$$(3^2+4^2)\cdot (x^2+y^2)\geq (3x+4y)^2\Rightarrow x^2+y^2\geq \frac{12^2}{5^2}\Rightarrow \sqrt{x^2+y^2}\geq \frac{12}{5}$$
So we get $$\min (z) = \min (\sqrt{x^2+y^2}) = \frac{12}{5}=\frac{a}{b}$$
So We get $$a+b = 12+5 = 17$$
A: Think of this geometrically. The distance between $3i$ and $4$ in the complex plane is $5$. The equation states that the length of the path from $3i$ to $z$ to $4$ equals the distance between $3i$ and $4$. This is possible only if $z$ is on the line segment between $3i$ and $4$ in the complex plane.
So find the point on that line segment that is closest to the origin (which is what it means to minimize $|z|$). Again, this is just geometry.
