# non zero complex numbers are orthogonal iff dot product is zero

Recall that the dot (scalar) product of two planar vectors $v_1 = (x_1, y_1)$ and $v_2 = (x_2 , y_2)$ is given by $v_1. v_2 = x_1x_2 + y_1y_2$.

Exercise: Show that the vectors represented by the (nonzero) complex numbers $z_1$ and $z_2$ are orthogonal if and only if $z_1. z_2 = 0$

Note/Recall: that the dot product of the vectors represented by the complex numbers $z_1$ and $z_2$ is given by $z_1. z_2$ = Re$(\bar {z_1} z_2)$.

And orthogonality holds precisely when $z_1 = icz_2$ for some real $c$.

Attempt: Suppose that the vectors represented by the (nonzero) complex numbers $z_1$ and $z_2$ are orthogonal. Then $z_1 = icz_2$ for some real $c$. Then $z_1. z_2 = icz_2(z_2) = ic|z_2|^2$

Converse: Suppose $z_1. z_2 = 0$, then $z_1 . z_2 = (x_1 + iy_1)(x_2 + iy_2) = x_1x_2 + ix_1y_2 + ix_2y_1 - y_1y_2 = x_1x_2 - y_1y_2 + i(x_1y_2 + x_2y_1)$ must be equal to zero?

So

• There would probably be less confusion if you said "orthogonal if and only if $\mathrm{Re}(\bar z_1 z_2) = 0.$" Then you can make a note afterward explaining that this is the complex dot product. (Also, try \cdot for the dots in the dot products.) – David K Feb 9 '15 at 5:15

You've defined $z_{1} \cdot z_{2} = Re(\overline{z_{1}} z_{2})$, and you want to show that $z_{1} \cdot z_{2} = 0 \iff z_{1} \perp z_{2}$. Without loss of generality, assume $\arg(z_{1}) \geq \arg(z_{2})$, and note that $z_{1} \perp z_{2}$ if, and only if, $\arg(z_{1}) - \arg(z_{2}) = \frac{\pi}{2}$ or $\arg(z_{1}) - \arg(z_{2}) = \frac{3\pi}{2}$.

Next, write the two complex numbers in polar form: $z_{1} = |z_{1}| e^{i \arg(z_{1})}$ and $z_{2} = |z_{2}| e^{i \arg(z_{2})}$. Then we have $$z_{1} \cdot z_{2} = |z_{1}||z_{2}| \cos(-\arg(z_{1}))\cos(\arg(z_{2}) - |z_{1}| |z_{2}| \sin(-\arg(z_{1})) \sin(\arg(z_{2})).$$ Because cosine is even and sine is odd, we have $$z_{1} \cdot z_{2} = |z_{1}||z_{2}| \cos(\arg(z_{1}))\cos(\arg(z_{2}) + |z_{1}| |z_{2}| \sin(\arg(z_{1})) \sin(\arg(z_{2})).$$ By trig identities, this is equal to $$z_{1} \cdot z_{2} = |z_{1}| |z_{2}| \cos(\arg(z_{1}) - \arg(z_{2})).$$ Since $|z_{1}| \neq 0$ and $|z_{2}| \neq 0$ (for, otherwise, $z_{1}$ and $z_{2}$ are trivially perpendicular), we must have $\cos(\arg(z_{1}) - \arg(z_{2})) = 0$, which can only happen if $\arg(z_{1}) - \arg(z_{2}) \in \{\frac{\pi}{2}, \frac{3\pi}{2}\}$.

For the converse, just work it backwards. We have $\arg(z_{1}) - \arg(z_{2})) \in \{\frac{\pi}{2}, \frac{3\pi}{2}\}$, so $z_{1} \cdot z_{2} = |z_{1}| |z_{2}| \cos( \arg(z_{1}) - \arg(z_{2})) = 0$.

So $z_{1} \cdot z_{2} = 0 \iff z_{1} \perp z_{2}$.

• Thank you for the help. On the first line, $z_1. z_2 = Re(z_1z_2)$, but inside inside $Re(z_1z_2)$ it's suppose to be the conjugate of $z_1$ times $z_2$ – user2842 Feb 9 '15 at 5:14
• You're right, that was a typo. In the rest of the argument, I used $z_{1} \cdot z_{2} = Re(\overline{z_{1}} z_{2})$. – Mark Feb 9 '15 at 5:15

In the first half of your proof, you either write $icz_2$ when you should have written $ic\bar z_2$, or you forgot to put a dot between the two $z_2$s.

In the second half, you forgot to conjugate $z_1$. You therefore end up with a couple of sign errors, but only one of them matters. Remember that only the real part (the part with the terms $x_1x_2$ and $y_1y_2$) needs to be zero; the part with a factor of $i$ can be anything.

Then remember that you are trying to prove that the vectors corresponding to these complex numbers are orthogonal. I assume you already know how to show that two vectors are orthogonal.

You might want to do your "converse" part first and see if you find an equation in it that can be made to do both the "if" and "only if" of the proof.