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Math Help - Complex Inequality

  1. #1
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    Complex Inequality

    Could anyone gimmie any clues with this please ?

    By writing z + 2 + i = (z + 2i) + (2-i)

    prove that  \sqrt {5} -2 \le |z + 2 + i| \le \sqrt {5} +2

    for z being in A where A= 1 \le |z+2i| \le 2
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by frodozoid View Post
    Could anyone gimmie any clues with this please ?

    By writing z + 2 + i = (z + 2i) + (2-i)

    prove that  \sqrt {5} -2 \le |z + 2 + i| \le \sqrt {5} +2

    for z being in A where A= 1 \le |z+2i| \le 2
    By the triangle inequality, \left|(z+2i)+(2-i)\right|\leqslant\left|z+2i\right|+\left|2-i\right|

    This then leads to the fact that the upper bound becomes \left|z+2+i\right|=\left|(z+2i)+(2-i)\right|\leqslant\left|z+2i\right|+\left|2-i\right|=2+\sqrt{(-1)^2+2^2}=\color{red}\boxed{2+\sqrt{5}}

    Now, for the lower bound, \left|\left|z+2i\right|-\left|2-i\right|\right|\leqslant\left|(z+2i)+(2-i)\right|=\left|z+2+i\right|. This implies that \left|\left|2-i\right|-\left|z-2i\right|\right|=\left|\sqrt{5}-2\right|={\color{red}\boxed{\sqrt{5}-2}}\leqslant\left|(z+2i)+(2-i)\right|=\left|z+2+i\right|

    Thus, {\color{red}\boxed{\sqrt{5}-2}}\leqslant\left|z+2+i\right|\leqslant{\color{red  }\boxed{\sqrt{5}+2}}
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  3. #3
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    Ah thanks Chris, I see what I had missed now.
    I was forgetting to take the magnitude of 2-i in the correct way, this is what was confusing me. That's where the root 5 comes from yeah.
    I had the first line written out but a stupid error like that and I was confused haha. Nice one man cheers !
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  4. #4
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    A geometric method: |z-a| can be interpreted as the distance between points z and a in the complex plane.

    Saying that z satisfies 1\le |(z+ 2i)- 0|\le 2 means that z+ 2i is in the "washer" with center 0, inner radius 1, outer radius 2.

    By "writing z+ 2+ i as z+ 2i+ 2-i" we can interpret |z+ 2+ i|= |(z+2i)- (i-2)| as the distance from the point z+2i to the point i-2. |i- 2|= \sqrt{5}> 2 so i- 2 is outside that "washer". The shortest distance from i- 2 to the larger circle is \sqrt{5}- 2 while the longest distance, to the other side of the outer circle, is \sqrt{5}+ 2.
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