# Math Help - Inverse of z

1. ## Inverse of z

Using $z^{-1} = \frac{z*}{|z|^2}$, geometrically show how you would construct $z^{-1}$.

NOTE z* means the conjugate of z.

I have no idea. I know how to show it algebraically, since we'd have:

$\frac{1}{(a+ib)} = \frac{(a-ib)}{a^2+b^2}$

Then, multiply the left side by the complex conjugate and we get the right hand side...

How'd I show this geometrically?

2. Plot the point z on a graph.
Reflect that point in the real axis; that is z conjugate.
Now divide by the absolute value of z squared.

3. Originally Posted by Plato
Plot the point z on a graph.
Reflect that point in the real axis; that is z conjugate.
Now divide by the absolute value of z squared.
How do you "divide" something on the graph...

So if we have our axis', we'd have:

ib on the vertical axis (above 0)
-ib on the vertical axis (below 0)

And we'd have a somewhere on the traditional x-axis (greater than 0)

So then a - ib will put us in the 4th quadrant, and that is our conjugate...

How do I graph a^2 + b^2 and then divide this by the point I already created from the conjugate?

4. Originally Posted by Ideasman
How do you "divide" something on the graph...

So if we have our axis', we'd have:

ib on the vertical axis (above 0)
-ib on the vertical axis (below 0)

And we'd have a somewhere on the traditional x-axis (greater than 0)

So then a - ib will put us in the 4th quadrant, and that is our conjugate...

How do I graph a^2 + b^2 and then divide this by the point I already created from the conjugate?
i suppose you divide the co-ordinates of the reflected point by the quantity a^2 + b^2

5. No idea how. I don't even know what my "b" is. It'd have to be "ib," no? A diagram would be helpful, but probably too hard to do on here.

6. Originally Posted by Ideasman
No idea how. I don't even know what my "b" is. It'd have to be "ib," no? A diagram would be helpful, but probably too hard to do on here.
well, plotting the complex number on an Argand diagram, you will get a point $(a,b)$. reflect in the real axis to obtain the point $(a,-b)$. then divide by the quantity $a^2 + b^2$ to get the point $\left( \frac a{a^2 + b^2}, - \frac b{a^2 + b^2}\right)$. this will be the geometric representation of the complex number $z^{-1}$

of course, this works if $a$ and $b$ are not both zero. if $a = b = 0$, then $z = z^{-1}$