Let $z$ be a complex number such that$|z-i|\leq5$, and let $z_1=5+3i$.
Find the minimum and maximum values of $|z_1+iz|$.
The geometric way to do this is easy, just draw a circle of radius $5$ centered at $(0,1)$ and find the minimum and maximum distances from there. But is there a way to do this purely algebraically?
My attempt:
Let $z=a+ib$
$\sqrt{a^2+(b-1)^2}\leq5$
$a^2+b^2-2b+1\leq25\qquad[1]$
Now, $|z_1+iz|=\sqrt{(5-b)^2+(3+a)^2}$
Adding $6a-8b+33$ to $[1]$, we get $|z_1+iz|^2\leq58+6a-8b$
I don't know where to go from here. Please help.
Best Answer
For a purely algebraic solution:
We have
$f(z) = |z_1+iz|^2 = (5-b)^2 + (3+a)^2 = 58 +6a - 8b = 50 + 6a - 8(b-1)$
It is clear that to maximise $f(z)$ subject to the constraint $a^2 + (b-1)^2 \le 25$, we must have $a^2 + (b-1)^2 = 25$, otherwise if $a^2 + (b-1)^2 < 25$ we could increase $a$ and/or decrease $b$ and so increase $f(z)$. So let $a=5 \sin \theta$ and $b-1 = 5 \cos \theta$. Then
$f(z) = 50 + 30 \sin \theta - 40 \cos \theta \\ \Rightarrow \frac {df}{d \theta} = 30 \cos \theta + 40 \sin \theta$
So $f$ has maximum and minimum values when
$30 \cos \theta + 40 \sin \theta = 0 \\ \Rightarrow \tan \theta = -\frac{3}{4} \\ \Rightarrow (\sin \theta, \cos \theta) = (\frac 3 5, - \frac 4 5) \text{ or } (- \frac 3 5, \frac 4 5)$
To maximise $f(z)$ we take the first pair of values, so
$f(z)_{max} = 50 + \frac {90} 5 + \frac {160} 5 = 100 \\ \Rightarrow |z_1+iz| = 10$