Solve {l}{ad^2+bc^2=0}{e=a^c}{text{Solvefor}ftext{where}}{f=e}

$$\left. \begin{array} { l } { a d ^ {2} + b c ^ {2} = 0 }\\ { e = a ^ {c} }\\ { \text{Solve for } f \text{ where} } \\ { f = e } \end{array} \right.$$

Solve for a, d, b, c, f (complex solution)

$$a=-\left(\frac{c}{d}\right)^{2}b\text{, }d=e^{-\frac{2\pi n_{1}iRe(c)}{\left(Re(c)\right)^{2}+\left(Im(c)\right)^{2}}-\frac{2\pi n_{1}Im(c)}{\left(Re(c)\right)^{2}+\left(Im(c)\right)^{2}}+\frac{Re(c)-iIm(c)}{\left(Re(c)\right)^{2}+\left(Im(c)\right)^{2}}}\text{, }n_{1}\in \mathrm{Z}\text{, }b\in \mathrm{C}\text{, }c\in \mathrm{C}\text{, }f=e\approx 2.718281828$$

$$a=e^{-\frac{2\pi n_{1}iRe(c)}{\left(Re(c)\right)^{2}+\left(Im(c)\right)^{2}}-\frac{2\pi n_{1}Im(c)}{\left(Re(c)\right)^{2}+\left(Im(c)\right)^{2}}+\frac{Re(c)-iIm(c)}{\left(Re(c)\right)^{2}+\left(Im(c)\right)^{2}}}\text{, }n_{1}\in \mathrm{Z}\text{, }d=0\text{, }b=0\text{, }c\in \mathrm{C}\text{, }f=e\approx 2.718281828$$

Solve for a, d, b, c, f

$$\left\{\begin{matrix}\\a=e^{\frac{1}{c}}\text{, }d=0\text{, }b=0\text{, }c\neq 0\text{, }f=e\approx 2.718281828\text{; }a=-\left(\frac{c}{d}\right)^{2}b\text{, }d=\sqrt{-b}ce^{-\frac{1}{2c}}\text{, }b<0\text{, }c\neq 0\text{, }f=e\approx 2.718281828\text{; }a=-\left(\frac{c}{d}\right)^{2}b\text{, }d=-\sqrt{-b}ce^{-\frac{1}{2c}}\text{, }b<0\text{, }c\neq 0\text{, }f=e\approx 2.718281828\text{, }&\text{unconditionally}\\a=-e^{\frac{1}{c}}\text{, }d=0\text{, }b=0\text{, }c\in \mathrm{R}\text{, }f=e\text{; }a=-\left(\frac{c}{d}\right)^{2}b\text{, }d=-\sqrt{b}ce^{-\frac{1}{2c}}\text{, }b>0\text{, }c\in \mathrm{R}\text{, }f=e\text{; }a=-\left(\frac{c}{d}\right)^{2}b\text{, }d=\sqrt{b}ce^{-\frac{1}{2c}}\text{, }b>0\text{, }c\in \mathrm{R}\text{, }f=e\text{, }&c\neq 0\text{ and }Denominator(c)\text{bmod}2=1\text{ and }Numerator(c)\text{bmod}2=0\end{matrix}\right.$$