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anonymous

  • 5 years ago

Use z = a + ib, w = c+id and the triangle inequality to prove that √[(a+c)² + (b+d)²] ≤ √(a² + b²) + √(c² + d²)

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  1. dumbcow
    • 5 years ago
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    square both sides and simplify \[ ac+bd \le \sqrt{a ^{2}+b ^{2}}\sqrt{c ^{2}+d ^{2}}\] square both sides and simplify \[2abcd \le b ^{2}c ^{2} + a ^{2}d ^{2}\] make substitution u =bc v = ad \[2uv \le u ^{2}+v ^{2}\] look at all 3 cases, u=v 2u^2 <= 2u^2 True u>v: u=v+1 \[2v(v+1)\le v ^{2}+(v+1)^{2}\] \[\rightarrow 2v ^{2}+2v \le 2v ^{2}+2v+1\] 0 <= 1 True u<v : u=v-1 \[\[2v(v-1)\le v ^{2}+(v-1)^{2}\] \[\rightarrow 2v ^{2}-2v \le 2v ^{2}-2v+1\] 0 <= 1 True

  2. anonymous
    • 5 years ago
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    Ignore the method above; it is not what they want. Consider |z| and |w|. We can consider these as the distances from the origin to each complex number in an argand diagram. Now consider |z - w|. What does this mean? This is: \[|z-w| = \sqrt{(a-c)^2 + (b-d)^2} \] so the distance between the two complex numbers, z and w. However, this is not going to be exactly what we want for the LHS, so let us instead consider the complex numbers z and -w (and |z-(-w)|. We work out the values of each of these things: \[ |z| = \sqrt{a^2 + b^2}\] \[ |-w| = \sqrt{(-c)^2 + (-d)^2} = \sqrt{c^2 + d^2}\] \[ |z-(-w)| = |z+w| = \sqrt{(a+c)^2 + (b+d)^2}\] But how can we use this to prove the inequality? Wait, DUH, we use the triangle inequality. Considering a triangle with sides a,b,c, we can say: \[a \leq b+c\] with the same holding for each side in a's position. How can this relate to our problem? We know |z| and |-w| are the distances from the origin to each of these, and |z+w| is the distance between them. OWAIT these form a triangle. From this we can say: \[|z+w| \leq |z| + |w| \] and the result follows immediately. ----- Sorry for the long post, the method is actually very quick but as you aren't here for me to guide through it I thought I'd explain everything - this method is FAR better than above (and what they actually wanted) Peace out, INewton.

  3. anonymous
    • 5 years ago
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    I'll be back later to collect my medals.

  4. anonymous
    • 5 years ago
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    lol

  5. anonymous
    • 5 years ago
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    here

  6. anonymous
    • 5 years ago
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    Unfortunately, you gave one to the first post too, so that is unacceptable.

  7. anonymous
    • 5 years ago
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    I thought you were harsh.

  8. anonymous
    • 5 years ago
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    Thank you very much! I thought that was a bit harsh too, but the first answer confused me. Yours was very helpful and the answer that I was looking for. Medal given ^^

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