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{array}{rcl}\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{array}

Now, recall that for any complex number z, z+\bar{z}=2\Re(z), and since \overline{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 z1z2 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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One Response to “Monday Math 147”

  1. Panayiota Katsamba Says:

    Hi!
    I really like your website, I think it’s really useful, but unfortunately, I cannot view any of the formulae, there’s a boxed cross.
    If you have any idea what the problem is, can you please let me know?

    Thanks,
    panayiota

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