Chapter 55

Complex Mappings55.1 Conformal Maps

If γ (t) = x(t)+ iy(t) is a C1 curve having values in U, an open set of C, and if f : U → Cis analytic, consider f ◦ γ, another C1 curve having values in C. Also, γ ′ (t) and ( f ◦ γ)′ (t)are complex numbers so these can be considered as vectors in R2 as follows. The complexnumber, x+iy corresponds to the vector, (x,y) . Suppose that γ and η are two such C1 curveshaving values in U and that γ (t0) = η (s0) = z and suppose that f : U→C is analytic. Whatcan be said about the angle between ( f ◦ γ)′ (t0) and ( f ◦η)′ (s0)? It turns out this angle isthe same as the angle between γ ′ (t0) and η ′ (s0) assuming that f ′ (z) ̸= 0. To see this, note(x,y) · (a,b) = 1

2 (zw+ zw) where z = x+ iy and w = a+ ib. Therefore, letting θ be thecosine between the two vectors, ( f ◦ γ)′ (t0) and ( f ◦η)′ (s0) , it follows from calculus that

cosθ

=( f ◦ γ)′ (t0) · ( f ◦η)′ (s0)∣∣( f ◦η)′ (s0)

∣∣ ∣∣( f ◦ γ)′ (t0)∣∣

=12

f ′ (γ (t0))γ ′ (t0) f ′ (η (s0))η ′ (s0)+ f ′ (γ (t0))γ ′ (t0) f ′ (η (s0))η ′ (s0)

| f ′ (γ (t0))| | f ′ (η (s0))|

=12

f ′ (z) f ′ (z)γ ′ (t0)η ′ (s0)+ f ′ (z) f ′ (z)γ ′ (t0)η ′ (s0)

| f ′ (z)| | f ′ (z)|

=12

γ ′ (t0)η ′ (s0)+η ′ (s0)γ ′ (t0)1

which equals the angle between the vectors, γ ′ (t0) and η ′ (t0) . Thus analytic mappings pre-serve angles at points where the derivative is nonzero. Such mappings are called isogonal..

Actually, they also preserve orientations. If z = x+ iy and w = a+ ib are two complexnumbers, then (x,y,0) and (a,b,0) are two vectors in R3. Recall that the cross product,(x,y,0)× (a,b,0) , yields a vector normal to the two given vectors such that the triple,(x,y,0) ,(a,b,0) , and (x,y,0)× (a,b,0) satisfies the right hand rule and has magnitudeequal to the product of the sine of the included angle times the product of the two normsof the vectors. In this case, the cross product will produce a vector which is a multiple ofk, the unit vector in the direction of the z axis. In fact, you can verify by computing bothsides that, letting z = x+ iy and w = a+ ib,

(x,y,0)× (a,b,0) = Re(ziw)k.

Therefore, in the above situation,

( f ◦ γ)′ (t0)× ( f ◦η)′ (s0)

= Re(

f ′ (γ (t0))γ′ (t0) i f ′ (η (s0))η ′ (s0)

)k

=∣∣ f ′ (z)∣∣2 Re

(γ′ (t0) iη ′ (s0)

)k

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