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Let $z_{1}, z_{2}, ..., z_{n}$ be nonzero complex numbers, with $z_{k}=p_{k}\exp(i\theta_{k})$, where $p_{k}$ is a positive real number and $\theta_{k}$ real. Can you help me prove that $\left | \sum_{k=1}^{n}z_{k} \right |^2=\sum_{k=1}^{n}(p_{k})^2+2\sum_{k<l}p_{k}p_{l}\cos(\theta_{k}-\theta_{l})$

Thank you

  • Could this help me establish an NSC for the modulus of the sum to be equal to the sum of the modulus ? –  Jan 11 '12 at 12:42

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Multiply: $$\left(\sum_j z_j\right)\left(\sum_k \bar{z}_k\right)=\sum_l|z_l|^2+\sum_{j\neq k}z_j\bar{z}_k.$$ In terms of your $p$ and $\theta$ we have $$ \sum_l p_l^2+\sum_{j\neq k} p_jp_ke^{i(\theta_j-\theta_k)}. $$ The cosine terms in $e^{i(\theta_j-\theta_k)}$ double up since cosine is even and the sine terms cancel since sine is odd, i.e. $$ \cos(\theta_j-\theta_k)+\cos(\theta_k-\theta_j)=2\cos(\theta_j-\theta_k) $$ and $$ \sin(\theta_j-\theta_k)+\sin(\theta_k-\theta_j)=0. $$ Hence you get the quoted result.

If you want $|\sum_l z_l|^2=(\sum_l |z_l|)^2$ then all of the $z_l$ must point in the same direction, in other words the $\theta_l$ must all be the same (draw a picture). Plugging that into what we have above, we get $$ \sum_l p_l^2+\sum_{j\neq k} p_jp_ke^{i(\theta_j-\theta_k)}=\sum_l p_l^2+\sum_{j\neq k}p_jp_k=\left(\sum_l p_l\right)^2. $$
yoyo
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  • Thank you for your proof. Can this help me establish a necessary and sufficient condition to have the modulus of the sum equal to the sum of the modulus' ? –  Jan 11 '12 at 10:33