Sorry for the late reply i was going to reply earlier but got side-tracked.

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Originally Posted by

**ThePerfectHacker** You did it the correct way, I just wanted you to get convinced that the map is angle perserving like I said. There happens to be an cheap way of doing this, that is to use the theorem: Mobius transformations map lines and circles into line and circles. Since $\displaystyle 1/z$ if a special case of the Mobius transformation it maps lines into circles or lines. So just pick three random points on the line and map them under $\displaystyle 1/z$ if they are collinear then you get a line, if not, then you get the unique circle which passes through those points. The proof of theorem that Mobius transformations map lines and circles into line and circles can be done with mindless algebra again, i.e. substitute $\displaystyle x+iy$ into $\displaystyle (az+b)/(cz+d)$ and get messy. But there is a nice way to proving it, I will post it later.

Yeah I figured out how to use my graphing packages parametric equation mode to get the graphs of the transformation, I don't fancy going through the mindless algebraic proof of this theorem. I'll be interested if you post it.

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If you remember I told you that many functions are conformal (angle preserving). There is a simple theorem in complex analysis which tells us which functions are conformal. If $\displaystyle f'(z_0) \not = 0$ for a point $\displaystyle z_0$ then it is conformal there. Meaning if $\displaystyle c_1,c_2$ are curves which intersect at $\displaystyle z_0$ then $\displaystyle f(c_1)$ and $\displaystyle f(c_2)$ are the same angle. Look how simple!

Seems almost too simple! What is the theorem called ? and how is it proved? If the proof is very detailed don't bother posting it, I can always purchase a book on the topic, could you recommend a good one though ?

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For example, $\displaystyle f(z) = e^z$ is conformal at any point. Because $\displaystyle f'(z_0) = e^{z_0} \not = 0$. Try it! Pick a few perpendicular lines emanating from the origin and find where they get mapped to. In particular try the easy ones first, i.e. $\displaystyle x=0$ and $\displaystyle y=0$ (which happen to be perpendicular).

this one is pretty easy (providing I got it right)

for $\displaystyle \Im(z) = 0$ $\displaystyle \Im(f(z)) = 0 \ \forall \ z \in \mathbb{C}$ and $\displaystyle \Re(z) > 0$ so it maps the line $\displaystyle y = 0$ onto the half line $\displaystyle \arg( \theta) = 0$

and for $\displaystyle x = 0$ then $\displaystyle f(z) = \sin y + i \cos y$ which is a circle in the complex plane and it intersects with the line at right angles at the point $\displaystyle (1,0)$

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Consider, $\displaystyle f(z) = z^2$, this function is conformal everywhere except possibly at $\displaystyle z_0 = 0$. Show that at $\displaystyle z_0 = 0$ the function is not conformal by finding a counterexample.

for $\displaystyle z = x + iy$ $\displaystyle f(z) = x^2 - y^2 + 2xyi$

so the line $\displaystyle x=0$ is mapped onto is mapped onto the half line $\displaystyle \arg(\theta) = \pi$ and the line $\displaystyle y=0$ is mapped onto the half line $\displaystyle \arg(\theta) = 0$. I am not sure if its correct to say that the lines either do not intersect at all at $\displaystyle z_0 = 0$ or if i say they intersect at the angle $\displaystyle \pi$, nevertheless I have show that function does not preserve angles at $\displaystyle z_0 = 0$.

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Finally, find where the integer grid (except the line $\displaystyle x=0$ and $\displaystyle y=0$ because they pass through the origin) gets mapped to. For instance, the line $\displaystyle x=1$ gets mapped into a parabola. In this problem, the algebra will get messy, so I suggest to use a graph program so graph these grid lines under $\displaystyle z^2$ and show they are orthogonal.

So the integer grid is comprised of the lines $\displaystyle x = \pm a \ \ \forall \ a \in \mathbb{N}$. so $\displaystyle f(z) = a^2 - y^2 \pm2iay$

so the parabolas are defined parametrically by $\displaystyle y = 2at$ and $\displaystyle x = a^2 -t^2$

lines $\displaystyle y = \pm a \ \ \forall \ a \in \mathbb{N}$

we have parabolas defined by $\displaystyle y = 2at$ and $\displaystyle x = t^2 - a^2$ so these parabola are reflection of the other ones in the y axis.

So the integer grid is transformed onto many intersection parabolas, I attached an image I made using my graphing package.

Now I must show that the tangents of these parabola intersect at right-angles.

for

$\displaystyle y = 2at$ and $\displaystyle x = a^2 -t^2$

$\displaystyle \frac{dy}{dx} = -\frac{a}{t}$

and for $\displaystyle y = 2bt$ and $\displaystyle x = t^2 - b^2$

$\displaystyle \frac{dy}{dx} = \frac{b}{t}$.

these lines before the intersection would have intersection at the point $\displaystyle (a,b)$ the new points are given by $\displaystyle t = b $ and $\displaystyle t =a$ respectively.

for the product of the gradients on the tangents of these parabola at these points is $\displaystyle -\frac{a}{b} \times \frac{b}{a} = -1$, hence they are perpendicular, so the function $\displaystyle f(z) = z^2$ is conformal on the interger grid, excluding the lines $\displaystyle y=0$ and $\displaystyle x=0$.

That was a brilliant exercise, many thanks TPH, sorry again for my late reply I really do enjoy and appreciate all your input.

best regards, Bobak