Complex numbers (1 Viewer)

[velox]

A complex number z has the property that when divided by (7 + 5i), its real part is twice its imaginary part. Find all possible solutions of z. Anyone got any ideas?

 

[McLake]

(a + bi)/(7 + 5i) = 2c + ci
(a +bi)(7 - 5i)/(49 + 25) = 2c + ci
(7a - 5ai + 7bi + 5b)/(74) = 2c + ci

so (7/74)a + (5/74)b = 2c
7a + 5b = 148c (1)
and (7/74)b - (5/74)a = c
7b - 5a = 74c (2)

so (1) * 5 = 35a + 25b = 740c
and (2) * 7 = 49b - 35a = 518c
so (1) + (2) = 74b = 1258c
b = 17c
sub back in (1),
7a + 5(17c) = 148c
a = 9c

so z = c(9 + 17i)

EDIT: dosn't have to be an int

 

[velox]

How do we get this [(7/74)b - (5/74)a = c] ?

 

[Trefoil]

(7a - 5ai + 7bi + 5b)/(74) = 2c + ci
gives an imaginary part of
(7b-5a)/74=c
hence
(7/74)b - (5/74)a = c

 

[velox]

oops, didnt know that. Kinda hard learning stuff with no textbook. Hmmm, must buy a textbook.

 

[SeDaTeD]

Alternatively,
arg( z/7+5i) = arctan(1/2)
argz = arg(7+5i) + arctan(1/2)
argz = arctan(5/7) + arctan(1/2)
tan(argz) = tan[ arctan(5/7) + arctan(1/2) ]
tan(argz) = [5/7 + 1/2]/[1 - 5/14]
tan(argz) = 17/9
argz = arctan(17/9)
therefore z= k(9+17i) where k is a real number