Monday Math 147

Given two non-zero complex numbers z₁ and z₂ such that $\left|z_1+z_2\right|=\left|z_1-z_2\right|$ , show that the arguments of z₁ and z₂ differ by π/2.

Recall that for any complex number z, $\left|z\right|^2=z\bar{z}$ , thus
$\begin{eqnarray}\left|z_1+z_2\right|^2&=&\left|z_1-z_2\right|^2\\\left(z_1+z_2\right)\left(\bar{z_1}+\bar{z_2}\right)&=&\left(z_1-z_2\right)\left(\bar{z_1}-\bar{z_2}\right)\\z_1\bar{z_1}+z_1\bar{z_2}+\bar{z_1}z_2+z_2\bar{z_2}&=&z_1\bar{z_1}-z_1\bar{z_2}-\bar{z_1}z_2+z_2\bar{z_2}\\z_1\bar{z_2}+\bar{z_1}z_2&=&-z_1\bar{z_2}-\bar{z_1}z_2\\z_1\bar{z_2}+\bar{z_1}z_2&=&{0}\end{eqnarray}$

Now, recall that for any complex number z, $z+\bar{z}=2\Re(z)$ , and since $\bar{z_1\bar{z_2}}=\bar{z_1}z_2$ , we see that
$z_1\bar{z_2}+\bar{z_1}z_2=2\Re\left(z_1\bar{z_2}\right)$ , and so
$\Re\left(z_1\bar{z_2}\right)=0$ ,
and we proved last week that z₁ and z₂ have arguments differing by π/2 (and thus give perpendicular vectors) if and only if $\Re\left(z_1\bar{z_2}\right)=0$ .

Consider the geometric interpretation. Given a parallelogram with sides given by the vectors corresponding to z₁ and z₂, then z₁+z₂ and z₁-z₂ correspond to the diagonals; thus, the above is equivalent to stating that if the diagonals of a parallelogram are of equal length, then the parallelogram is a rectangle.

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Tags: Complex Numbers, Geometry, Math, Monday Math

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Monday Math 147

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