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 - x^{2}, 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.

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 - x^{2}, 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 $\displaystyle b<0$, then $\displaystyle g(x)$ is no longer one-to-one. $\displaystyle g(-x) = 1-(-x)^2 = 1-x^2 = g(x)$.

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.

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: $\displaystyle \dfrac{z+i}{z+2} = i$ implies $\displaystyle z+i = (z+2)i$. Plug in $\displaystyle z = a+bi$, multiply out, and equate coefficients.

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 - x

^{2}, 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 $\displaystyle \displaystyle \begin{align*} z = a + i\,b \end{align*}$ at all...

$\displaystyle \displaystyle \begin{align*} \frac{z + i}{z + 2} &= i \\ z + 1 &= i\left( z + 2 \right) \\ z + 1 &= i\,z + 2i \\ z - i\,z &= -1 + 2i \\ z \left( 1 - i \right) &= -1 + 2i \\ z &= \frac{-1 + 2i}{1 - i} \\ z &= \frac{ \left( -1 + 2i \right) \left( 1 + i \right) }{ \left( 1 - i \right) \left( 1 + i \right) } \\ z &= \frac{ -1 - i + 2i - 2}{1^2 + 1^2} \\ z &= -\frac{3}{2} + \frac{1}{2}i \end{align*}$

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.

Re: Inverse Functions & Complex Numbers