Precalculus method to prove $ab^*+cd^*=0$

Question

Question. Suppose $a,b,c,d\in\mathbb C$ and the following equations holds: $$|a|^2+|b|^2=|c|^2+|d|^2=1$$ $$ac^*+bd^*=0$$ where $x^*$ is the complex conjugate of $x$. Prove that $ab^*+cd^*=0$.

I already have a solution involving some basic knowledge of linear algebra:
Denote $X=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$. It is a unitary matrix since $$XX^H=\left(\begin{matrix}aa^*+bb^*&ac^*+bd^*\\ca^*+db^*&cc^*+dd^*\end{matrix}\right)=\left(\begin{matrix}|a|^2+|b|^2&ac^*+bd^*\\(ac^*+bd^*)^*&|c|^2+|d|^2\end{matrix}\right)=I_2$$ Therefore, $$I_2=X^HX=\left(\begin{matrix}-&-\\ab^*+cd^*&-\end{matrix}\right)$$Hence the result.
Motivation
Because this question looks elementary, out of curiosity, I'm wondering if there is a method without using matrices.
I tried letting $(a,b,c,d)=(e^{i\alpha}\sin\varphi,e^{i\beta}\cos\varphi,e^{i\gamma}\sin\theta,e^{i\delta}\cos\theta)$ then plug it into the second known relation but I don't know how to simplify it.

Answer 1

We know that $$\begin{cases} aa^*+bb^*=1\\ cc^*+dd^*=1\\ ac^*+bd^*=0 \end{cases}$$

Starting from the third equality we multiply it with $b^*c$ to get

$$ab^*c^*c+bb^*cd^*=0$$ now using the first two we get $$ab^*(1-dd^*)+cd^*(1-aa^*)=0$$ so $$ab^*+cd^*=ab^*dd^*+aa^*cd^*$$ and inserting the third again this time conjugated ($a^*c+b^*d=0$) in the RHS we get $$ab^*+cd^*=ab^*dd^*-ab^*dd^*=0$$ which is exactly what we were looking for.

Answer 2

For convenience, let $A=a^*$, $B=b^*$, $C=c^*$, and $D=d^*$. We then have $$aA+bB=1,$$ $$cC+dD=1,$$ $$aC+bD=0,$$ and $$Ac+Bd=0.$$ First we want to show that $aA=dD$. This follows from \begin{align}dD&=dD\cdot 1=dD(aA+bB)=adAD+bdBD\\&=adAD+(bD)(Bd)=adAD+(-aC)(-Ac)\\&=adAD+acAC=aA(cC+dD)=aA\cdot 1=aA.\end{align}

If $d=0$, then $aA=dD=0$ so that $a=0$. Hence $aB+cD=0B+c0=0$. We now assume $d\ne 0$. Therefore \begin{align}aB+cD&=aB+c\left(\frac{dD}{d}\right)=aB+c\left(\frac{aA}{d}\right)\\&=\frac{a}{d}\left(Bd+Ac\right)=\frac{a}{d}\left(Ac+Bd\right)=\frac{a}{d}\cdot 0=0.\end{align}

From this result, it follows that $$(a,b,c,d)=\big(e^{i\alpha}\cos \vartheta,e^{i\beta}\sin\vartheta,e^{i\gamma}\sin\vartheta,-e^{i(-\alpha+\beta+\gamma)}\cos\vartheta),$$ where $\alpha,\beta,\gamma\in[0,2\pi)$ and $\vartheta\in[0,\pi/2]$.