Let z be a complex such that:

1<=|z|<=6.

Then what are the minimum and maximum values for the complex (z+3).

I suspect that:

0<=|z+3|<=9, but I can't find a way to prove it.

Please help!

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- Mar 15th 2008, 06:40 AMtombrowningtonmodulus of a comples
Let z be a complex such that:

1<=|z|<=6.

Then what are the minimum and maximum values for the complex (z+3).

I suspect that:

0<=|z+3|<=9, but I can't find a way to prove it.

Please help! - Mar 15th 2008, 06:57 AMMoo
Hello,

I don't understand well, are you asked to find the minimum and maximum values for the complex (z+3) or the minimum and maximum values for the modulo of (z+3) ? - Mar 15th 2008, 08:40 AMtombrownington
Hi Moo,

Sorry I didn't put it well.

It's the maximum value for the modulus of (z+3) that I was looking for. - Mar 15th 2008, 08:54 AMMoo
I'll try, but i find the question weird :

Replace z by a+ib.

Hence $\displaystyle |z| = \sqrt{a^2+b^2}$

So $\displaystyle 1 \leq \sqrt{a^2+b^2} \leq 6$

$\displaystyle 1 \leq a^2+b^2 \leq 36$ (squared the previous inequations)

z+3 would be (a+3)+ib

=> $\displaystyle |z+3| = \sqrt{(a+3)^2+b^2} = \sqrt{a^2 + b^2 + 6a + 9} = \sqrt{|z|^2 + 6a + 9}$

So $\displaystyle \sqrt{1+6a+9} \leq |z+3| \leq \sqrt{36+6a+9}$

$\displaystyle \sqrt{6a+10} \leq |z+3| \leq \sqrt{45+6a}$

And that's what i find strange : there is no information about a :( - Mar 15th 2008, 09:02 AMtombrownington
Moo, I already tried that method and came to the same result.

I thought of it in a different way - the proof isn't at all rigorous; in fact it's just an insight not a proof.

By adding 3 to z we are shifting the point representing z (on the complex plane) 3 units to the right.

Now since z can be -3, this gives us |z+3|=0. So this has to be the minimum - the smallest possible value for a modulus being 0.

Also if z = 6, then |z+3| = 9, and you will find that whatever other point z you take, the modulus |z+3| will always be smaller than 9.

But as I said, this is only an insight. I can't get any further.

Do you see something that I don't? - Mar 15th 2008, 09:16 AMMoo
Well, i need to think again about it. There must be a way to put a restriction to a and to make it fit with your supposition (which seems correct ! ;))

- Mar 15th 2008, 09:50 AMwingless
This is what the problem wants to say:

Find such |z| that $\displaystyle 1 \leq |z| \leq 6$ and that maximizes (or minimizes) |z+3|.

$\displaystyle z = x + yi$

$\displaystyle |z| = \sqrt{x^2 + y^2}$

$\displaystyle z+3 = x + 3 + yi$

$\displaystyle |z+3| = \sqrt{(x+3)^2 + y^2}$

$\displaystyle 1 \leq \sqrt{x^2 + y^2} \leq 6$

$\displaystyle 1 \leq x^2 + y^2 \leq 36$

$\displaystyle |z+3| = \sqrt{(x+3)^2 + y^2}$

Now this is only an optimization question =)

$\displaystyle \frac{d}{dx}|z+3| = \frac{3+x}{\sqrt{(3+x)^2+y^2}}$

$\displaystyle \frac{3+x}{\sqrt{(3+x)^2+y^2}} = 0$

$\displaystyle x=-3$

$\displaystyle |z+3| = \sqrt{(-3+3)^2 + y^2}$

$\displaystyle |z+3| = \sqrt{y^2}$

$\displaystyle y=0$ will make $\displaystyle |z+3|$ minimum. So,

$\displaystyle |z+3| = \sqrt{(-3+3)^2 + 0^2} = \boxed{0}$

For maximum,

The maximum x and y values that satisfy $\displaystyle 1 \leq x^2 + y^2 \leq 36$ and maximize $\displaystyle \sqrt{(x+3)^2 + y^2}$ are 6 and 0.

This is how to find it:

$\displaystyle 1 \leq x^2 + y^2 \leq 36$

Maximize: $\displaystyle \sqrt{(x+3)^2 + y^2}$

Maximize: $\displaystyle \sqrt{\underbrace{x^2 + y^2}_{\text{Maximum: 36}} + 6x + 9}$

There is no term of y in 6x + 9 so x must be maximum. Then $\displaystyle x=6$ and $\displaystyle y=0$

$\displaystyle |z+3| = \sqrt{(x+3)^2 + y^2}$

$\displaystyle |z+3| = \sqrt{(6+3)^2 + 0^2} = \boxed{9}$ - Mar 15th 2008, 09:57 AMwingless
- Mar 15th 2008, 04:49 PMThePerfectHacker
Note, $\displaystyle |z+3|\leq |z|+|3| \leq 6+3=9$. Thus, $\displaystyle |z+3|$ is bounded by $\displaystyle 9$ if $\displaystyle 1\leq |z|\leq 6$. To show this is the best possible bound (and therefore maximum) just note that $\displaystyle 1\leq |6|\leq 6$ and then $\displaystyle |z+3| = 9\leq 9$.

To show the minimum of $\displaystyle |z+3|$ is zero is to just note that $\displaystyle |z+3|\geq 0$. Now we claim this is the best lower bound, and therefore minimum. If $\displaystyle z=-3$ then $\displaystyle 1\leq |z|\leq 6$ and $\displaystyle |z+3| = 0$.