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If $|z-i|\leq 2,z_0 = 5+3i$ then find max value of $|iz+z_0|.$

Now ,

$|iz+z_0| \leq |iz|+|z_0| \cdots \text{(By Triangle Inequality.)}$

Now consider the diagram in Argand Plane - enter image description here

Observe that $|z| \leq 3.$

$\Rightarrow |iz+z_0| \leq \boxed{3 + \sqrt{34}}$

But the Answer is $\boxed{7}.$ What is wrong in this proof $?$

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  • $\begingroup$ Can anyone please help? $\endgroup$
    – Aleph
    Commented Nov 21, 2022 at 15:15

4 Answers 4

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Where you were incorrect is not accounting for the change of $z \to iz$. While its true that $|z| = |iz|$, this does not always apply when adding complex numbers, $|z + z_0| ≠ |iz + z_0|$.

In your method, when you substituted $|iz| = 3$, you assimilated that similar substitution also applies to $|iz + z_0|$ towards, as you would for $|z + z_0|$.

If $z = x + yi$, then $|z + z_0| = |x + 5 + (y + 3)i|$; its easy to see that you would choose a set of possible positive ${x, y}$ within the bound $|z - i| \le 2$, while obtaining the longest length ($|z + z_0|_{max}$).

However, $|iz + z_0| = |xi - y + 3i + 5|$. Notice the $y$ is now negated. This means that using positive $y$ values, will reduce the coeff of the complex number $iz + z_0$ (as $x_{z0} > 0$ and $y_{z0} > 0$) and decrease $|iz + z_0|$, instead of obtaining the maximum.

Instead, you need a set of ${x, -y}$ to obtain $|iz + z_0|_{max}$. [$x$ is unchanged as rotation is only by $\frac{\pi}{2}$, $z\to iz$].

Graphically, your solution lies in 4 quadrant (sector) of the $|z - i| = 2$:

Sector

While I did not yet solve it algebraically, I tried trial and error and reached the approximation of $\approx 7$

Approximation

Graphical Point

Finding an exact answer is left as an exercise.


UPDATE

While the the triangle inequality is not used in the method below, consider if it might suit the question better.

From above, we know the solution lies on the arc of the curve, $y = -\sqrt{4 - x^2} + 1, x \in [0, \sqrt{3}]$

Hence, substituting this for $x, y$ in the complex number $iz$, we get:

$$|iz + z_0|$$ $$\to |ix - y + 5 + 3i|$$ $$\to |ix - \big[-\sqrt{4 - x^2} + 1 \big] + 5 + 3i|$$ $$\to |(x + 3)i + (\sqrt{4 - x^2} + 4)|$$

$$\to \sqrt{(x + 3)^2 + \big[ \sqrt{4 - x^2} + 4 \big]^2}$$

$$\to \sqrt{8\sqrt{4 - x^2} + 6x + 29}$$

In order to find the maximum length, we need to find where the length function above reaches a maximum.

Hence,

$$\frac{d}{dx}\sqrt{8\sqrt{4 - x^2} + 6x + 29} = 0$$

$$\frac{d}{dx}\big[8\sqrt{4 - x^2} + 6x + 29\big] \cdot \frac{1}{\sqrt{8\sqrt{4 - x^2} + 6x + 29}} = 0$$

We have numerator = 0,

$$\to 3\sqrt{4 - x^2} + 4x = 0$$ $$\therefore x = \frac{6}{5}, x \in [0, \sqrt{3}]$$

Finally,

$$\to |(\frac{6}{5} + 3)i + (\sqrt{4 - \big(\frac{-6}{5}\big)^2} + 4)|$$ $$ = 7 \text{, as required}$$

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The question can be solved as follows:

Note that

$(1)$ $|z-i| \leq 2$ represents a circle centered at $i$ with radius $2$.

$(2)$ \begin{align} |iz+z_o| & = |i|\cdot|iz+z_o| \\ & = |i^2z+iz_o| \\ & = |iz_o-z| \end{align}

This can be interpreted as the distance between $z$ and $iz_0$.

$(3)$ $iz_0=i(5+3i)=-3+5i$

Thus the question can be interpreted as finding the maximum distance between a point inside or on the circle $|z-i| \leq 2$ from a fixed point $iz_o=-3+5i$

enter image description here

From the figure, it is obvious that the greatest distance is attained at point $D$.

Note that $Z_0C=|iz_0-i|=|(-3+5i)-i|=|-3+4i|=5$

The greatest distance is $5+2=7$

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While everything you've done is correct, the fact that

$$\lvert iz+z_0\rvert\leq3+\sqrt{34}$$

does not mean that $3+\sqrt{34}$ is the maximum. So while you have found an upper bound, it it not necessarily the smallest upper bound. If it was, then there would be a $z$ for which it was attained, which there is not. Instead, notice that with

$$\lvert z-i\rvert\leq 2$$

and

$$z_0=5+3i$$

we have that

$$\lvert iz+z_0\rvert=\lvert iz+5+3i\rvert=\lvert i(z-5i+3)\rvert=\lvert z-i+3-4i\rvert\leq\lvert z-i\rvert+\lvert 3-4i\rvert\leq 2+5=7.$$

To show then that this is actually the maximum, you need to find some $z$ with $\lvert z-i\rvert\leq2$ such that $\lvert iz+z_0\rvert=7$. I'll leave that to you as I don't quite have the time for it right not, but you should be able to find it either by just playing around with the possible values, or perhaps by parameterizing the circle $\lvert z-i\rvert =2$ and trying to optimize using calculus techniques somehow (if you know calculus that is, but you should be able to find it regardless).

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Hint:

$$|iz+5+3i| = |z - 5i + 3|$$

Now use the given condition and ∆ inequality.


As for the error in your proof, it's due to using the upper bound $|z|\le3$ and not the upper bound $|z-i|\le2$. The latter is a subset of the former. $|z|\le3$ is not true for all $z$ in the disk $|z-i|\le2$.

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