Inverse of complex numbers...help! stuck on function f(a+ib)=(1-2a)+i(3b+2)

[gordonhayward]

I've been stuck on this for a while now.
for the function:
f(a+ib)=(1-2a)+i(3b+2)

could someone please spell out the steps. I truly have no idea. Thanks

 

[pka]

I've been stuck on this for a while now.
for the function: f(a+ib)=(1-2a)+i(3b+2) could someone please spell out the steps. I truly have no idea.

Well I too am clueless as to what the question is?
Here is an example: \(\displaystyle f(3-7{\bf{i}})=(1-2(3)+{\bf{i}}(3(-7)+2)=-5-19\bf{i}\)
BUT your post is titled 'inverse of complex numbers'. Inverse?? What inverse?

What are you asking?

 

[Dr.Peterson]

Are you asking how to find the inverse of this function?

One way is to suppose that f(a+ib) = x+iy, that is, (1-2a)+i(3b+2) = x+iy. Then you need to solve for a and b in terms of x and y.

What is the first step? What must be true if two complex numbers are equal?