1. modulus 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.

2. 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) ?

3. 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.

4. I'll try, but i find the question weird :

Replace z by a+ib.

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

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

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

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

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

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

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

And that's what i find strange : there is no information about a

5. 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?

6. 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 ! )

7. This is what the problem wants to say:

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

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

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

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

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

Now this is only an optimization question =)

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

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

$x=-3$

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

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

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

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

For maximum,

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

This is how to find it:
$1 \leq x^2 + y^2 \leq 36$

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

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

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

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

$|z+3| = \sqrt{(6+3)^2 + 0^2} = \boxed{9}$

8. Originally Posted by tombrownington
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.
Note, $|z+3|\leq |z|+|3| \leq 6+3=9$. Thus, $|z+3|$ is bounded by $9$ if $1\leq |z|\leq 6$. To show this is the best possible bound (and therefore maximum) just note that $1\leq |6|\leq 6$ and then $|z+3| = 9\leq 9$.

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