• # question_answer Solving $3-2yi={{9}^{x}}-7i$, where ${{i}^{2}}=-1,$ for x and y real, we get [AMU 2000] A) $x=0.5\,\,,\,\,y=3.5$ B) $x=5\,\,,\,\,y=3$ C) $x=\frac{1}{2}\,\,,\,\,y=7$ D) $x=0,\,y=\frac{3+7i}{2i}$

$3-2yi={{9}^{x}}-7i$ Equating real and imaginary parts both sides ${{9}^{x}}=3\Rightarrow \,{{3}^{2x}}={{3}^{1}}\Rightarrow 2x=1\Rightarrow x=0.5$ $2y=7\Rightarrow \,y=3.5$.