z_{i} are complex numbers. Show that |Re[ (z_{1}^{2} + z_{2}^{2} + ... + z_{n}^{2})^{1/2} ]| ≤ |Re z_{1}| + |Re z_{2}| + ... + |Re z_{n}|.

**Solution**

Let z_{k} = x_{k} + i y_{k}, and let (z_{1}^{2} + z_{2}^{2} + ... + z_{n}^{2})^{1/2} = a + ib.

Then, squaring, a^{2} - b^{2} = ∑ x_{k}^{2} - ∑ y_{k}^{2} (1), ab = ∑ x_{k}y_{k} (2). The Cauchy inequality gives that |∑ x_{k}y_{k}| ≤ ( ∑ x_{k}^{2} )^{1/2} ( ∑ y_{k}^{2} )^{1/2}, so from (1) we have |ab| ≤ ( ∑ x_{k}^{2} )^{1/2} ( ∑ y_{k}^{2} )^{1/2} (3). If |a| > ( ∑ x_{k}^{2} )^{1/2}, then from (3) |b| < ( ∑ y_{k}^{2} )^{1/2}. But then a^{2} - b^{2} > ∑ x_{k}^{2} - ∑ y_{k}^{2}, contradicting (1). So we must have that |a| ≤ ( ∑ x_{k}^{2} )^{1/2}. The required result follows immediately since obviously ( ∑ |x_{k}| )^{2} ≥ ∑ x_{k}^{2}.

© John Scholes

jscholes@kalva.demon.co.uk

4 Dec 1999