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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?

gueesstt  Apr 22, 2018

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 #1
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\((5+3i)(a+bi)=32-8i\\ 5a+5bi+3ai-3b=32-8i\\ 5a-3b+(5b+3a)i=32-8i\\ \)

\(so\\ 5a-3b=32 \qquad and \qquad 3a+5b=-8 \\ 25a-15b=160 \quad and \qquad 9a+15b=-24 \\ add\\ 34a=136\\ 17a=68\\ a=\frac{68}{17}\\ a=4 \)

 

\(3a+5b=-8\\ 3*4+5b=-8\\ 5b=-20\\ b=-4\\ \)

Ramanujan pick          \(4-4i\)

Melody  Apr 22, 2018
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 #1
avatar+92458 
+3
Best Answer

\((5+3i)(a+bi)=32-8i\\ 5a+5bi+3ai-3b=32-8i\\ 5a-3b+(5b+3a)i=32-8i\\ \)

\(so\\ 5a-3b=32 \qquad and \qquad 3a+5b=-8 \\ 25a-15b=160 \quad and \qquad 9a+15b=-24 \\ add\\ 34a=136\\ 17a=68\\ a=\frac{68}{17}\\ a=4 \)

 

\(3a+5b=-8\\ 3*4+5b=-8\\ 5b=-20\\ b=-4\\ \)

Ramanujan pick          \(4-4i\)

Melody  Apr 22, 2018

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