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Math Help - modulus of a comples

  1. #1
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    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.

    Please help!
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  2. #2
    Moo
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    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) ?
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  3. #3
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    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.
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  4. #4
    Moo
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    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
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  5. #5
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    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?
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  6. #6
    Moo
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    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 ! )
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  7. #7
    Super Member wingless's Avatar
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    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}
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    Super Member wingless's Avatar
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  9. #9
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    Quote Originally Posted by tombrownington View Post
    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.
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