# Bijection in a complex number ring

1. Mar 12, 2009

### dancergirlie

1. The problem statement, all variables and given/known data
Show that the complex conjugation function f:C----->C (whose rule is f(a+bi)=a-bi) is a bijection

2. Relevant equations
A function is a bijection if it is both injective and surjective
a function is injective if when f(a)=f(b) then a=b
a function is surjective if for f:B---->C
if c is in C then there exists a b in B such that f(b)=c

3. The attempt at a solution
I get the whole injective part
Let
f(a+bi)=f(c+di)
hence, a-bi=c-di
now suppose that b is unequal to d
so, a-c=(b-d)i
meaning i=(a-c)/(b-d) which is a contradiction, since i is not a real number.
hence, a=c and b=d
meaning, f is injective

My problem is with the surjective part, let me know if this is wrong,
Let f(a+bi)=c+di
then, a-bi=c+di.... then i don't know where to go from there
I don't know if this is wrong, any clarification on how to test for surjectivity would be greatly appreciated...

2. Mar 12, 2009

### Staff: Mentor

You can prove this by contradiction, but why? If two complex numbers are equal, what must be true for their real parts and their imaginary parts?
What you want to do is to show that for any complex number z2 in the range of this function, there is a complex number z1 in its domain so that f(z1) = z2. Start with an arbitrary complex number and find the other complex number that this function maps to it.

3. Mar 12, 2009

### sutupidmath

Like Mark suggested.

Let z1=a-bi be any complex number, then for this exists its conjugate z2=a+bi such that
f(z2)=z1.