# An algebraic way to do this question? Find minimum and maximum values of $|z_1+iz|$ where $|z-i|\leq5,\:\:z_1=5+3i$.

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 $$z=a+ib$$

$$\sqrt{a^2+(b-1)^2}\leq5$$

$$a^2+b^2-2b+1\leq25\qquad$$

Now, $$|z_1+iz|=\sqrt{(5-b)^2+(3+a)^2}$$

Adding $$6a-8b+33$$ to $$$$, we get $$|z_1+iz|^2\leq58+6a-8b$$

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

$$f(z)_{max} = 50 + \frac {90} 5 + \frac {160} 5 = 100 \\ \Rightarrow |z_1+iz| = 10$$

• Could you provide some intuition for the fact that the maximisation of $f$ happens at equality? Apr 5, 2020 at 11:50
• Suppose $a^2+(b-1)^2 < 25$ - say we take $a=4$ and $b=-0$ so that $a^2 + (b-1)^2 = 17$ and $f(z)=82$. Notice that if we increase $a$ or decrease $b$ then we increase $f(z)$. Since we don't have equality in the constraint on $a$ and $b$, we can increase $a$ to, say, $4.5$ which gives $f(z)=85$. Or we can decrease $b$ to, say, $-1$ which gives $f(z)=90$. So the maximum value of $f(z)$ must occur at a point where we cannot increase $a$ or decrease $b$, which means we must be at the boundary of the feasible region and $a^2+(b-1)^2 = 25$. Apr 6, 2020 at 11:12

Note that $$|z_1+iz|=\big|i(z_1+iz)\big|=|iz_1-z|$$. From the triangle ineq, we get $$5\ge |z-i|=\big|(iz_1-i)-(iz_1-z)\big|\geq \big||iz_1-i|-|iz_1-z|\big|=\big||z_1-1|-|z_1+iz|\big|.$$ Therefore $$0=5-|z_1-1|\le |z_1+iz|\le 5+|z_1-1|=10.$$ The lhs is an equality iff $$z=iz_1=-3+5i$$. The rhs is an equality iff $$z=3-3i$$.

This technique works in a more general setting. Let $$z_0$$ and $$z_1$$ be complex numbers. For a given $$r>0$$, let $$C$$ be the circular region given by $$|z-z_0|\le r$$. Then, $$|z_1-z_0|-r\leq |z_1-z_0|-|z-z_0|\leq |z-z_1|\leq |z_1-z_0|+|z-z_0|\leq |z_1-z_0|+r.$$ The maximum value $$|z_1-z_0|+r$$ occurs iff $$z=z_0+r\left(\frac{z_1-z_0}{|z_1-z_0|}\right)$$ in the case $$z_1\ne z_0$$, and $$z$$ is on the boundary of $$C$$ if $$z_1=z_0$$. For the minimum value, when $$|z_1-z_0|\ge r$$, then the minimum value is $$r-|z_1-z_0|$$, which occurs iff $$z=z_0-r\left(\frac{z_1-z_0}{|z_1-z_0|}\right)$$ in the case $$z_1\ne z_0$$, and $$z$$ is on the boundary of $$C$$ if $$z_1=z_0$$. If $$|z_1-z_0|, then the minimum value is $$0$$, which is achieved when $$z=z_1$$.

You can just use geometry. Complex numbers $$z$$ s.t. $$|z-i|\le 5$$ form an area inside and on the boundary of the circle with centre $$(0,1)$$ and radius $$5$$. This region contains the complex number $$iz_1$$. Therefore the minimum value of $$|z_1+iz|=|iz_1-z|$$ is $$0$$. The maximum value of $$|z_1+iz|=|iz_1-z|$$ is the diameter of the circle, which is $$10$$.

Hint: it is beneficial to do some rewriting first. Denote $$w=iz+z_1$$ then $$|z-i|=|iz+1|=|w+1-z_1|\le 5.$$ Now call $$c=z_1-1=4+3i$$ to end up with $$\min/\max |w|\quad\text{subject to }|w-c|\le 5=|c|.$$

As opposed to $$z=a+ib$$, it is more convenient to use the polar form for the algebraic solution. Let $$z-i=re^{i\theta}, \> r\in[0,5]$$ and $$iz_1-i=3-4i=5e^{i\theta_0}$$. Then,

$$|z_1+i z| = |i z_1 -z|=|5e^{i\theta_0}-re^{i\theta} | = \sqrt{ 25 -10r\cos( \theta-\theta_0) +r^2}$$

Since $$-1\le \cos( \theta-\theta_0) \le 1$$, we have

$$0\le 5-5 \le 5-r \le |z_1+i z| \le 5+r \le 5+5=10$$

Let $$z$$ be lying on the disc $$S:|z-i|\leq 5$$ \begin{aligned} d:=\left|z_{1}+i z\right| =\left|z-z_{1} i\right|=|z-(-3+5 i)|, \end{aligned} which is equal to the distance of $$z$$ from the complex number $$-3-5i$$.

$$d$$ attains its maximum and minimum values at $$z_M$$and $$z_m$$ on the disc $$S$$ respectively iff $$z_1, z_m$$ and $$z_M$$ are collinear such that

\begin{aligned} d_{\min } &=\left|z_{m}-(3+5 i)\right| \\ &=|i+3-5 i|-5 \\ &=\sqrt{3^{2}+4^{2}}-5 \\ &=0 \end{aligned}\begin{aligned} d_{\max } &=\left|z_{M}-(-3+5 i)\right| \\ &=|i+3-5 i|+5 \\ &=5+5 \\ &=10 \end{aligned}

Remark: $$iz_1=-3+5i$$ is on the circle simply implies the maximum and minimum $$d$$ are respectively $$0$$ and $$10$$. The above method can be used when $$iz_1$$ lies outside the circle.