728 CHAPTER 36. ISOLATED SINGULARITIES AND ANALYTIC FUNCTIONS

g′g is an analytic function on B(z0,r) and so by Morera’s theorem, Theorem 35.6.1, it

has a primitive on B(z0,r) called h. Therefore by the product rule and the chain rule,(ge−h

)′= g′

(e−h)+g(−e−h

)h′

= g′(

e−h)+g(−e−h

) g′

g= 0

and so there exists a constant, C = ea+ib such that on B(z0,r) ,

ge−h = ea+ib.

Therefore,g(z) = eh(z)+a+ib

and so, modifying h by adding in the constant, a+ ib it is still a primitive of g′/g and nowg(z) = eh(z) where h′ (z) = g′(z)

g(z) on B(z0,r) . Letting

φ (z) = (z− z0)eh(z)m

implies formula 36.1 is valid on B(z0,r) . Now φ (z0) = 0 but

φ′ (z0) = e

h(z0)m ̸= 0.

Shrinking r if necessary you can assume φ′ (z) ̸= 0 on B(z0,r). Is there an open set V

contained in B(z0,r) such that φ maps V onto B(0,δ ) for some δ > 0?Let φ (z) = u(x,y)+ iv(x,y) where z = x+ iy. Consider the mapping(

xy

)→

(u(x,y)v(x,y)

)

where u,v are C1 because φ is given to be analytic. The Jacobian of this map at (x,y) ∈B(z0,r) is ∣∣∣∣∣ ux (x,y) uy (x,y)

vx (x,y) vy (x,y)

∣∣∣∣∣=∣∣∣∣∣ ux (x,y) −vx (x,y)

vx (x,y) ux (x,y)

∣∣∣∣∣= ux (x,y)

2 + vx (x,y)2 =

∣∣φ ′ (z)∣∣2 ̸= 0.

This follows from a use of the Cauchy Riemann equations. Also(u(x0,y0)

v(x0,y0)

)=

(00

)

Therefore, by the inverse function theorem there exists an open set V, containing z0 andδ > 0 such that (u,v)T maps V one to one onto B(0,δ ) . Thus φ is one to one onto B(0,δ )as claimed. Applying the same argument to other points z of V and using the fact thatφ′ (z) ̸= 0 at these points, it follows φ maps open sets to open sets. In other words, φ

−1 iscontinuous.

728 CHAPTER 36. ISOLATED SINGULARITIES AND ANALYTIC FUNCTIONSe is an analytic function on B(zo,r) and so by Morera’s theorem, Theorem 35.6.1, ithas a primitive on B(zo,r) called h. Therefore by the product rule and the chain rule,(6) = eC) ene(er)ee-en) eaand so there exists a constant, C = e*+’? such that on B(zo,r),ge! _ eatibTherefore,h(z)-+a+ibgz) =eand so, modifying / by adding in the constant, a+ ib it is still a primitive of g'/g and nowg(z) =e" where h’ (z) = a on B(zo,r). Letting6(2)=(e-m)emimplies formula 36.1 is valid on B(zo,r). Now @ (zo) = 0 but6! (@) =e ” 40.Shrinking r if necessary you can assume @’(z) £0 on B(zo,r). Is there an open set Vcontained in B(zo,r) such that @ maps V onto B(0,6) for some 6 > 0?Let @ (z) =u(x,y) +iv (x,y) where z =x + iy. Consider the mappingx \_( u(y)y v(x,y)where u,v are C! because @ is given to be analytic. The Jacobian of this map at (x,y) €B(z0,r) isVe(X,y) Ux (x,y)Ux (x,y) uy (x,y)Vx (x,y) Vy (x,y)Ux (x,y) Vx (x,y) |= ux (x,y)? +x (x,y)? = |! (2)|? 40.This follows from a use of the Cauchy Riemann equations. Alsou(xo,yo) \_ ( 0v(x0,Y0) 0Therefore, by the inverse function theorem there exists an open set V, containing zy and5 > 0 such that (u,v)’ maps V one to one onto B(0,5). Thus @ is one to one onto B(0, 8)as claimed. Applying the same argument to other points z of V and using the fact that@' (z) £0 at these points, it follows @ maps open sets to open sets. In other words, @~! iscontinuous.