Isnt it Max or min?

Since the angle at $z$ sub tended by $z_1$ and $z_2$ is $\tfrac14\pi$ , $z$ must lie on the portion of the circle centre $7+9i$ and radius $3\sqrt{2}$ that lies above the line $\mathrm{Im}\,z =6$ . The centre of this circle subtends an angle $\tfrac12\pi$ at $z_1$ and $z_2$ . The desired distance is $3\sqrt{2}$ , so the answer is $2(2+3) = \boxed{10}$ .

$\large \displaystyle \arg\left ( \frac{z - z_1}{z - z_2} \right ) = \frac{\pi}{4}$

$\large \displaystyle \arg(z - z_1) - \arg(z - z_2) = \frac{\pi}{4}$

Let $z = x + yi$ :

$\large \displaystyle \tan^{-1} \left (\frac{y-6}{x-10} \right ) - \tan^{-1} \left (\frac{y-6}{x-4} \right ) = \frac{\pi}{4}$

$\large \displaystyle \tan \left [ \tan^{-1} \left (\frac{y-6}{x-10} \right ) - \tan^{-1} \left (\frac{y-6}{x-4} \right ) \right ] = \tan \left ( \frac{\pi}{4} \right )$

$\large \displaystyle \frac{ \frac{y-6}{x-10} - \frac{y-6}{x-4}}{1 + \frac{ (y-6)^2 }{(x-4)(x-10)}} = 1$

$\large \displaystyle \frac{(y-6)(x-4) - (y-6)(x-10)}{(x-10)(x-4)} = \frac{(x-10)(x-4) + (y-6)^2}{(x-10)(x-4)}$

$\color{#20A900} \boxed{\large \displaystyle -y^2 + 18y - 72 = x^2 - 14x + 40 }$

$\large \displaystyle | z - 7 - 9i| = \sqrt{ (x-7)^2 + (y-9)^2 }$

$\large \displaystyle | z - 7 - 9i| = \sqrt{ x^2 - 14x + 49 + y^2 - 18y + 81 }$

$\large \displaystyle | z - 7 - 9i| = \sqrt{ x^2 - 14x + 40 + 9 + y^2 - 18y + 81 }$

$\large \displaystyle | z - 7 - 9i| = \sqrt{ -y^2 + 18y - 72 + 9 + y^2 - 18y + 81 }$

$\large \displaystyle | z - 7 - 9i| = \sqrt{18}$

$\color{#20A900} \boxed{ \large \displaystyle | z - 7 - 9i| = 3 \sqrt{2} }$

$\large \displaystyle \color{#3D99F6} a = 3, b = 2, \boxed{ \large \displaystyle 2(a+b) = 10 }$