# Inverse Functions & Complex Numbers

• November 19th 2013, 10:22 PM
Fratricide
Inverse Functions & Complex Numbers
1.http://i.imgur.com/VdOUGtn.png
(Sorry if this is a little big.)

I don't really know where to start. I substituted z = a + bi into the equation, but that didn't seem to help me.

2. Let g:[b, 2] --> R, where g(x) = 1 - x2, if b is the smallest real value such that g has an inverse function, find b and g-1(x).

I know that b = 0, but why? I can find the inverse function, but I can't quite grasp why 0 is the smallest value which allows g to have an inverse function.
• November 19th 2013, 10:59 PM
SlipEternal
Re: Inverse Functions & Complex Numbers
Quote:

Originally Posted by Fratricide
1.http://i.imgur.com/VdOUGtn.png
(Sorry if this is a little big.)

I don't really know where to start. I substituted z = a + bi into the equation, but that didn't seem to help me.

Try it again. That is the correct method.

Quote:

Originally Posted by Fratricide
2. Let g:[b, 2] --> R, where g(x) = 1 - x2, if b is the smallest real value such that g has an inverse function, find b and g-1(x).

I know that b = 0, but why? I can find the inverse function, but I can't quite grasp why 0 is the smallest value which allows g to have an inverse function.

Because if $b<0$, then $g(x)$ is no longer one-to-one. $g(-x) = 1-(-x)^2 = 1-x^2 = g(x)$.
• November 19th 2013, 11:14 PM
Melody2
Re: Inverse Functions & Complex Numbers
hi Fratricide
I will offer half an appology to start with because you probably know a lot more than me
but
with the first one couldn't you just substitute in the value of z and then multiply top and bottom by the conjugate of the denominator
then put the real part=0 and the coefficient of i = 1 ?
I did it and I think it worked but I might have made a mistake.
Melody.
• November 19th 2013, 11:29 PM
SlipEternal
Re: Inverse Functions & Complex Numbers
Quote:

Originally Posted by Melody2
hi Fratricide
I will offer half an appology to start with because you probably know a lot more than me
but
with the first one couldn't you just substitute in the value of z and then multiply top and bottom by the conjugate of the denominator
then put the real part=0 and the coefficient of i = 1 ?
I did it and I think it worked but I might have made a mistake.
Melody.

Another method: $\dfrac{z+i}{z+2} = i$ implies $z+i = (z+2)i$. Plug in $z = a+bi$, multiply out, and equate coefficients.
• November 19th 2013, 11:52 PM
Prove It
Re: Inverse Functions & Complex Numbers
Quote:

Originally Posted by Fratricide
1.http://i.imgur.com/VdOUGtn.png
(Sorry if this is a little big.)

I don't really know where to start. I substituted z = a + bi into the equation, but that didn't seem to help me.

2. Let g:[b, 2] --> R, where g(x) = 1 - x2, if b is the smallest real value such that g has an inverse function, find b and g-1(x).

I know that b = 0, but why? I can find the inverse function, but I can't quite grasp why 0 is the smallest value which allows g to have an inverse function.

There's no need to plug in \displaystyle \begin{align*} z = a + i\,b \end{align*} at all...

\displaystyle \begin{align*} \frac{z + i}{z + 2} &= i \\ z + 1 &= i\left( z + 2 \right) \\ z + 1 &= i\,z + 2i \\ z - i\,z &= -1 + 2i \\ z \left( 1 - i \right) &= -1 + 2i \\ z &= \frac{-1 + 2i}{1 - i} \\ z &= \frac{ \left( -1 + 2i \right) \left( 1 + i \right) }{ \left( 1 - i \right) \left( 1 + i \right) } \\ z &= \frac{ -1 - i + 2i - 2}{1^2 + 1^2} \\ z &= -\frac{3}{2} + \frac{1}{2}i \end{align*}
• November 19th 2013, 11:53 PM
Melody2
Re: Inverse Functions & Complex Numbers
Hi SlipEternal
Thanks, that makes sense to me. Only trouble is I get 2 different answers. I'll take a break and then try to see where my stupid mistake lies.
Melody.
• November 20th 2013, 01:04 PM
Fratricide
Re: Inverse Functions & Complex Numbers
Thanks for the help.