# Complex functions with a real variable (graphs)

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1. Feb 14, 2017

### Poetria

1. The problem statement, all variables and given/known data

How do the values of the following functions move in the complex plane when t (a positive real number) goes to positive infinity?

y=t^2

y=1+i*t^2

y=(2+3*i)/t

3. The attempt at a solution

I thought:

y=t^2 - along a part of a line that does not pass through the origin

y=1+i*t^2 - along a part of parabola

y=(2+3*i)/t - along a part of hyperbola

Unfortunately everything is wrong. I understand that e.g. y=1+i*t^2 is a line =1 and a parabola but I don't know how to connect it. Could you give me a hint how to visualise this?
Other possibilities: spirals inward/outward, clockwise/counterclockwise, along a circle, radially inward/outward

2. Feb 14, 2017

### PeroK

With $t$ as a parameter, you are not plotting $y$ against $t$, but simply the locus of $y$ in the plane.

3. Feb 14, 2017

### Poetria

I am trying to imagine it. So e.g. t^2 would move counterclockwise along a circle? (Moving parabola?)

4. Feb 14, 2017

### Poetria

I got the second one right, moving along part of a line... Wow! I am beginning to understand.

5. Feb 14, 2017

### PeroK

Note that for the first locus $y$ is always real.

6. Feb 14, 2017

### Poetria

Is it simply a parabola in this case?

7. Feb 14, 2017

### Poetria

As for the third I guess it spirals clockwise inward as t^(-1).

8. Feb 14, 2017

### PeroK

Why not simply plot $y$? You seem to be still thinking that you are plotting real $y$ against real $t$ on a normal 2D graph. That's not the case at all.

You are plotting a single complex number $y$ as its value changes.

If $y$ is real then it is confined to the real line and cannot be a parabola!

9. Feb 14, 2017

### PeroK

No. In particular, I'm not sure how you got the clockwise motion?

10. Feb 14, 2017

### LCKurtz

Perhaps there would be less confusion for the OP if the variable had been called $z$ instead of $y$. Especially since complex numbers are usually expressed as $z = x + iy$. Pretty poor notation for the problem if you ask me.

11. Feb 14, 2017

### Poetria

Thank you very much for your patience. :)
I tried plotting and got the third right.

12. Feb 14, 2017

### Poetria

Ok. I got all the three right. Phew. It is very simple in fact. :(