# How Relate Complex and Cartesian Functions For $\frac{1}{1+z^2}$, $\frac{1}{1-x^2}$, and $\frac{1}{1+x^2}$

This question is clumsily phrased: please ignore it and instead see the improved version here.

The graph of $$\frac{1}{1+z^2}$$ is:

The graph of $$\frac{1}{1+x^2}$$ is:

And the graph of $$\frac{1}{1-x^2}$$ Is:

And the graph of $$\left|\frac{1}{1-x^2}\right|$$ Is:

Notice how the graph of $\frac{1}{1+x^2}$ is equivalent to cutting the complex graph with a vertical plane directly between the two singularities (along the real axis). Notice also how cutting the complex graph with a vertical plane throug the two singualrites (perpendicular to the real axis) yeilds $|\frac{1}{1-x^2}|$. Obviously the complex plane and the cartesian plane bear some relation that it somewhat baffeling:

1.) Why must $\frac{1}{1-x^2}$ have it's absolute value taken before matching the complex function $\frac{1}{1+z^2}$?

2.) Graphing $\frac{1}{1-z^2}$ instead of $\frac{1}{1+z^2}$ rotates the singularities from $0,\pm i$ to $\pm 1, 0i$ or rotates the complex into the real and the real into the complex, but the same cartesian graphs apply. This suggests that what is on the real axis and what is on the imaginary axis is arbitrary because in one case the graph $\frac{1}{1+x^2}$ corresponds to the the vertical slice along the real axis and in the other case it corresponds to the vertical slice along the imaginary axis. The point of saying it is arbitrary is that $\frac{1}{1+x^2}$ is no more real or imaginary than $\frac{1}{1-x^2}$. Is that a valid conclusion?

3.) Is it possible to write a cartesian equation equivalent to a vertical plane through the complex plane $\frac{1}{1+z^2}$ for every possible angle and not just perpendicular or parallel to the real axis? If yes, how?

AntonioVargas shared the following plot of $Re(1/(1+z^2))$ colored according to $Im(1/(1+z^2))$:

This plot explains why the Absolute value was necessary for the Cartesian function to represent the Complex function: The complex function had its absolute value taken as well. Questions 2 & 3 remain open.

[edit 2]

Thanks to pbs for finally explaining how to graph a complex function! I was able to graph the Re(1/(1+z^2) on Wolfram Alpha!

However the 3D plot of the imaginary component begs more questions than it answers:

It would be so much simpler to explain with out the Imaginary plot! What's going on here? The absolute value seems to have been taken without having been stated. Hmmmm... Back to question 1 as unanswered! (except this time only for $Im(1/(1+z))$.

• Think about what you mean by 'the graph of' a holomorphic function.
– Rhys
Commented Aug 7, 2013 at 15:04
• The height of your graph of $1/(1+z^2)$ is $|1/(1+z^2)|$ and it is colored according to $\arg 1/(1+z^2)$. Commented Aug 7, 2013 at 15:09
• @AntonioVargas Is there a way to graph it 3D without the absolute values? Commented Aug 7, 2013 at 22:43
• You could plot $\operatorname{Re}\Bigl(1/(1+z^2)\Bigr)$ and color it according to $\operatorname{Im}\Bigl(1/(1+z^2)\Bigr)$ as in this image. Commented Aug 7, 2013 at 23:36
• The answers to this question have revealed to me that this question is clumily phrased and so I am asking a cleaned up version here. Commented Aug 8, 2013 at 22:40

You can understand complex functions better by expanding in terms of real variables. For example, expand $$\frac{1}{1+z^2}$$ by letting $z=x+iy$. Then $$\frac{1}{1+(x+iy)^2}=\frac{1}{1+x^2-y^2+2xyi}=\frac{1+x^2-y^2-2xyi}{(1+x^2-y^2)^2+4x^2y^2}.$$ You can then separate to obtain $$\frac{1}{1+z^2}=\frac{1+x^2-y^2}{(1+x^2-y^2)^2+4x^2y^2}-i\frac{2xy}{(1+x^2-y^2)^2+4x^2y^2}.$$ Investigating this will help you to compare the complex function with your alternative real function more precisely.

• I'm a little confused what you did between $1/1+x^2...$ and $1+x^2..../()^2+4x^2 y^2$. Commented Aug 8, 2013 at 8:27
• Multiply the numerator and denominator by the conjugate of the denominator, e.g. if the fraction were $$\frac{1}{u+iv}$$ then you would multiply numerator and denominator by $u-iv$, thus $$\frac{u-iv}{(u+iv)(u-iv)}.$$ This is perfectly valid since we are just multiplying by $1$ in "disguise". We call $u-iv$ the complex conjugate of $u+iv$. This is the method we use to make a complex denominator of a fraction purely real. Commented Aug 8, 2013 at 8:28
• Thankyou soooo much for explaining how to graph complex function!!!!! I have tried for a lllloooonnnggg time to figure that out! Everyone is like "buy Maple/Mathematica" or quit. If you wouldn't mind copying and pasting the answer over here I will gladly make it the accepted answer there. Commented Aug 8, 2013 at 8:41
• You're welcome. Commented Aug 10, 2013 at 14:06

The singluarities of $\frac{1}{1+z^2}$ are at plus/minus i...

Consider $z = (1 + c)i, e > 0$

Then,

$$\frac{1}{1+z^2} = \frac{1}{1 - (1 + c)^2} < 0$$

So you're first graph is wrong... the graph should become negative near the singularities

• Welcome to MSE! It really helps readability to format answers using MathJax (see FAQ). Regards Commented Aug 7, 2013 at 17:03
• Technically it is the graph of $\frac{1}{1-z^2}$, but I don't have maple or mathematica and Sage only does 2D graphs. Commented Aug 7, 2013 at 21:26