if z and w are complex numbers , prove that | |z| - |w| | <= |z-w| Solution Sinc

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Explanation / Answer

Since z and w are complex numbers, we can write them as: z = a + bi and w = c + di, where a, b, c, and d are some real numbers. By definition: |z| = v(a^2 + b^2) and |w| = v(c^2 + d^2), so: |z|^2 = a^2 + b^2 and |w|^2 = c^2 + d^2 ==> 2(|z|^2 + |w|^2) = 2a^2 + 2b^2 + 2c^2 + d^2. Now, we have: (i) z + w = (a + bi) + (c + di) = (a + c) + (b + d)i (ii) z - w = (a + bi) - (c + di) = (a - c) + (b - di). So: |z + w|^2 = (a + c)^2 + (b + d)^2 and |z - w|^2 = (a - c)^2 + (b - d)^2. Adding these up give: |z + w|^2 + |z - w|^2 = (a + c)^2 + (b + d)^2 + (a - c)^2 + (b - d)^2 = (a^2 + 2ac + c^2) + (b^2 + 2bd + d^2) + (a^2 - 2ac + c^2) + (b^2 - 2bd + d^2) = 2a^2 + 2b^2 + 2c^2 + 2d^2, which is precisely 2(|z|^2 + |w|^2). Therefore: 2(|z|^2 + |w|^2) = |z + w|^2 + |z - w|^2, as required.

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