+0

# Help?

+1
418
1
+603

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?

Apr 22, 2018

#1
+101761
+3

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

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

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

Ramanujan pick          $$4-4i$$

.
Apr 22, 2018

#1
+101761
+3

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

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

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

Ramanujan pick          $$4-4i$$

Melody Apr 22, 2018