+610 Ramanujan and Hardy played a game where they both picked a complex number. If the product of their numbers was $32-8i$, and Hardy picked $5+3i$, what number did Ramanujan pick?
+118609 \((5+3i)(a+bi)=32-8i\\ 5a+5bi+3ai-3b=32-8i\\ 5a-3b+(5b+3a)i=32-8i\\ \)
\(so\\ 5a-3b=32 \qquad and \qquad 3a+5b=-8 \\ 25a-15b=160 \quad and \qquad 9a+15b=-24 \\ add\\ 34a=136\\ 17a=68\\ a=\frac{68}{17}\\ a=4 \)
\(3a+5b=-8\\ 3*4+5b=-8\\ 5b=-20\\ b=-4\\ \)
Ramanujan pick \(4-4i\)
+118609 \((5+3i)(a+bi)=32-8i\\ 5a+5bi+3ai-3b=32-8i\\ 5a-3b+(5b+3a)i=32-8i\\ \)
\(so\\ 5a-3b=32 \qquad and \qquad 3a+5b=-8 \\ 25a-15b=160 \quad and \qquad 9a+15b=-24 \\ add\\ 34a=136\\ 17a=68\\ a=\frac{68}{17}\\ a=4 \)
\(3a+5b=-8\\ 3*4+5b=-8\\ 5b=-20\\ b=-4\\ \)
Ramanujan pick \(4-4i\)
