We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
236
1
avatar+169 

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?

 Jul 2, 2018
 #1
avatar
+1

Simplify the following:
(-8 i + 32)/(3 i + 5)

Factor 8 out of 32 - 8 i giving 8 (4 - i):
(8 (-i + 4))/(3 i + 5)

Multiply numerator and denominator of (8 (-i + 4))/(3 i + 5) by 5 - 3 i:
(8 (-i + 4) (-3 i + 5))/((3 i + 5) (-3 i + 5))

(5 + 3 i) (5 - 3 i) = 5×5 + 5 (-3 i) + 3 i×5 + 3 i (-3 i) = 25 - 15 i + 15 i + 9 = 34:
(8 (-i + 4) (-3 i + 5))/34

The gcd of 8 and 34 is 2, so (8 (-i + 4) (-3 i + 5))/34 = ((2×4) (-i + 4) (-3 i + 5))/(2×17) = 2/2×(4 (-i + 4) (-3 i + 5))/17 = (4 (-i + 4) (-3 i + 5))/17:
(4 (-i + 4) (-3 i + 5))/17

(4 - i) (5 - 3 i) = 4×5 + 4 (-3 i) - i×5 - i (-3 i) = 20 - 12 i - 5 i - 3 = 17 - 17 i:
(4 -17 i + 17)/17

Factor 17 out of 17 - 17 i giving 17 (1 - i):
(4×17 (-i + 1))/17
Combine powers. (4×17 (-i + 1))/17 = 4×17^(1 - 1) (-i + 1):
4×17^(1 - 1) (-i + 1)

1 - 1 = 0:
4×17^0 (-i + 1)

17^0 = 1:
4 (-i + 1) = 4 - 4i - Ramanujan's number.

 Jul 2, 2018

14 Online Users