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!
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
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?
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 $\displaystyle y=0$
$\displaystyle |z+3| = \sqrt{(x+3)^2 + y^2}$
$\displaystyle |z+3| = \sqrt{(6+3)^2 + 0^2} = \boxed{9}$
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 $\displaystyle |z+3| = 0$.