Imaginary to Imaginary and Real to Real Divisions? No Way!

Assume there exists at least a solution of $(a,b,c,d)$ that satisfies the following equation

$\begin{aligned}\dfrac{a + bi}{c + di}&= \dfrac{a}{c} + \dfrac{b}{d}i\\\\ \implies a+bi&=(c+di)\times \left(\dfrac{a}{c} +\dfrac{b}{d}i\right)\\ &=(a-b)+\left(\dfrac{bc}{d}+\dfrac{ad}{c}\right)i\\ \text{Equating} &\text{ real parts we get,}\\ a&=(a-b)\\ \implies b&=0\hspace{5mm}\color{#3D99F6}\small(1)\\ \text{Equating} &\text{ imaginary parts we get,}\\ b&=\left(\dfrac{bc}{d}+\dfrac{ad}{c}\right)\\ 0&=\dfrac{ad}{c}\hspace{5mm}\color{#3D99F6}\small\text{from}(1)\\ \text{since }d\neq0 &\text{ we must have }a=0\\ \implies (a+bi)&=0+0i\\ \text{Since } a+bi &\text{ is a nonzero complex number, our assumption that a solution exists is wrong.}\\ \text{Thus there} &\text{ exists no solution for the given equation.}\end{aligned}$

Let's rationalize the first quotient of complex numbers according to:

$\frac{a+bi}{c+di} \cdot \frac{c-di}{c-di} = \frac{(a+bi)(c-di)}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} + \frac{bc-ad}{c^2+d^2} i$ (i).

We now require matching the real and the imaginary parts above:

$\frac{ac + bd}{c^2 + d^2} = \frac{a}{c}$ & $\frac{bc-ad}{c^2+d^2} = \frac{b}{d}$ ;

or $bc = ad$ (REAL) & $bcd - ad^{2} = bc^2 + bd^2$ (IMAGINARY) (ii). Substituting the real expression into the imaginary side simplifies the latter into:

$bcd - ad^{2} = bc^2 + bd^2 \Rightarrow (ad)d - ad^2 = (ad)c + bd^2 \Rightarrow ad^2 - ad^2 = d(ac + bd) \Rightarrow 0 = d(ac + bd)$ (iii)

Since $d \ne 0$ , we require $ac + bd = 0 \Rightarrow d = -\frac{ac}{b}$ (iv). A final substitution back into the real expression of (ii) now yields:

$bc = ad \Rightarrow bc = a (-\frac{ac}{b}) \Rightarrow b^{2}c = -a^{2}c \Rightarrow a^2 + b^2 = 0 \Rightarrow a = b = 0$

which makes the original complex number $a + bi = 0$ (CONTRADICTION).