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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.

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5 Answers 5

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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

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

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  • $\begingroup$ Could you provide some intuition for the fact that the maximisation of $f$ happens at equality? $\endgroup$ Commented Apr 5, 2020 at 11:50
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    $\begingroup$ 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 $a^2+(b-1)^2 = 25$. $\endgroup$
    – gandalf61
    Commented Apr 6, 2020 at 11:12
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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 $|z_1-z_0|<r$, 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 $10$.

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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 $$ \min/\max |w|\quad\text{subject to }|w-c|\le 5=|c|. $$

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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

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

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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.

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