II.4 Inverse Mappings and the Jacobian 1

Inverse Mappings and the Jacobian

1. Sketch the gradient vector fields ∇ u and ∇ v for (a) uiv=e z , (b) uiv=Log z.

(a) e z=exiy=ex e iy=ex cos y i [e x sin y] . Thus, ∇ u=⟨ex cos y ,−ex sin y ⟩ and∇ v=⟨e x sin y ,e x cos y⟩ . Whence, the vector field looks like

(b) Log z=log∣z∣i Arg z . Whence, u=1/ 2 log x2 y2 and

v={0 if x=0 and y=0arctan y / x if x≥0arctan y / x if x0

. Thus the vector field looks like

II.4 Inverse Mappings and the Jacobian 2

2. Let a be a complex number, a≠0 , and let f z be an analytic branch of z a on ℂ∖ (−∞ , 0 ] .Show that f ' z =a f z / z.

Without loss of generality, suppose f z =ea Log z e2 i a m. Then, by the chain rule, we getf ' z =e2 i am ea Log z⋅a⋅1/ z=a f z / z.

3. Consider the branch f z = z 1−z on ℂ∖[0,1] that has positive imaginary part at z = 2. What is f ' z ? Be sure to specify the branch for the expression for f ' z .

On the principal branch, −2=∣−2∣1/2e i / 2 =2cos /2isin /2=2⋅i is positive. Now,

f ' z = z−z21/2=1/2 z−z 2−1 /2⋅1−2 z = 1−2 z2 z 1−z

. (c.f. Q2.)

4. Find the derivative of Tan−1 z. Find the derivative of tan−1 z for any analytic branch of the function defined on a domain D.

We would expect that ddz

Tan−1 z = ddz [ 1

2 iLog1iz

1−iz ]. Using properties of the principal

branch,this equalsddz [ 1

2 iLog 1iz−Log1−iz]= 1

2 i i1i z

i1−iz = 1

z21. By the definition of

log, we know that the derivative will be the same for any branch. (As the only difference between any two branches is a constant which vanishes after differentiation.)

5. Suppose g(z) is an analytic branch of cos−1 z , defined on a domain D. Find g ' z . Do different branches of cos−1 z have the same derivative?

ddz

cos−1 z =i⋅ ddz

[Log z± z2−12i m]=i⋅1±1/2 z 2−1−1 /2⋅2 z z± z2−1

= ±i z 2−1

. Thus, there are

two possible derivatives. (Again, the constant term vanishes after differentiation.)

6. Suppose h(z) is an analytic branch of sin−1 z , defined on a domain D. Find h' z . Do different branches of sin−1 z have the same derivative?

Since sin−1 z :=−i log iz±1−z 2, we can see a priori that there will be two different derivatives.

7. Let f (z) be a bounded analytic function, defined on a bounded domain D in the complex plane, and suppose that f (z) is one-to-one. Show that the area of f (D) is given by

Area f D=∬D∣ f ' z∣2 dxdy .

First, Area f D =∬ f Ddx dy . From the change of variable formula, we can compose the constant

function with f to obtain ∬DDet J f dxdy=∬D

∣ f ' z ∣dx dy. This completes the proof.

II.4 Inverse Mappings and the Jacobian 3

8. Sketch the image of the circle {∣z−1∣≤1} under the map w= z2. Compute the area of the image.

The circle has its center at z = 1 and has radius 1. Under the map, the radius will be squared. However, this image is the very same circle. Thus, the area is .

9. Compute∬D

∣ f ' z ∣2 dx dyfor f z =z 2 and D the open unit disk {∣z∣1}. Interpret your answer in terms of area.

From Q7, the integral is , the area of the circle under the image.

10. Show that if z= f is a one-to-one analytic function from a bounded domain V onto U, thenDU g ,h=DV g ° f ,h° f .

By definition, DU g , h=∬U [∂ g∂ x

∂ h∂ x

∂ g∂ y

∂ h∂ y ]dx dy. Now, U= f V , so we obtain

DU g , h=∬V [ ∂g° f ∂ x

∂h° f ∂ x

∂g ° f ∂ y

∂h° f ∂ y ]=DV g ° f ,h° f , as desired.