Let $a$ be a complex number and $b$ be imaginary show $ab$ is orthogonal to $a$

Question

Let $a$ be a complex number of the form $x+y \sqrt{d}$ where $x$ and $y$ are integers and $d$ is a negative integer. And let $b$ be an imaginary number equal to $\sqrt{d}$. Prove that $a$ and $ab$ are orthogonal.

Is there an orthogonal test for complex numbers? I tried to make a parallelogram with vertices $a,ab, a+ab,0$ in the complex plane and show that it is a rectangle by calculating the length of the diagonals but failed.

Can someone help me please?

We basically have to show $(x+y\sqrt{d})\sqrt{d})+x+y\sqrt{d}$ has equal norm to $(x+y\sqrt{d})\sqrt{d}-x-y\sqrt{d}$. Or just prove $ab$ and $a$ are orthogonal

(My book assumes this was trivial)

Answer 1

Mutiplying a complex number $z$ by $i$, you get $iz$, which is what you get if you rotate $z$ around the origin by a $90^\circ$ degrees clockwise rotation. So, $z$ and $iz$ are orthogonal. And $z$ and $-iz$ are orthogonal too, since $-iz$ is what you get if you rotate $z$ around the origin by a $90^\circ$ degrees anti-clockwise rotation.

If $b$ is a square root of $d$ with $d<0$, then $b=\pm\sqrt{-d}\,i$ and therefore $a$ and $ab$ are orthogonal.

Answer 2

Think of complex numbers as $2$-dimensional vectors. You want the dot product of such two vectors to be $0$, if they are to be orthogonal. Now we can find $ab=(x+y\sqrt{d})(\sqrt{d})=x\sqrt{d}+yd = yd + ix\sqrt{-d}$. In vector form this would be \begin{bmatrix} yd\\ x\sqrt{-d}. \end{bmatrix} You also know that $a$ can be represented by the vector \begin{bmatrix} x\\ y\sqrt{-d}. \end{bmatrix} Taking the dot product of the two gives $xyd - xyd = 0$.

Answer 3

You can write any complex number as $z=\vert z\vert e^{i \arg(z)}$, which are basically the polar coordinates consisting of the angle $\arg(z)$ and length. Multiplication by $i = e^{i\frac{\pi}{2}}$ is thus just multiplication by 90 degrees. In other words, $z$ and $iz$ are orthogonal. In your case $\sqrt{ d} = \sqrt{-\vert d\vert} = i\sqrt{\vert d\vert}$ acts like a multiplication by 90 degrees (via the multiplication by $i$) and a scaling by the factor $\sqrt{\vert d\vert}$, which does not touch orthogonality...