LECTURE 2: COMPLEX DIFFERENTIATION AND CAUCHY

1 LECTURE 2: COMPLEX DIFFERENTIATION AND CAUCHYRIEMANN EQUATIONSWe have seen in the first LECTURE that the COMPLEX derivative of a functionfat apointz0is defined as the limitf (z0) = limh 0f(z0+h) f(z0)h,whenever the limit exist. We have also seen two examplesi) iff(z) =z2thenf (z) = 2z, ii) the functionf(z) =zis not a differentiable function. Now we willgo for a detail differentiable atz0thenfis continuous (z0) = limz z0f(z) f(z0)z z0it follows thatlimz z0f(z) = limz z0f(z) f(z0)z z0(z z0) +f(z0) =f(z0).

2 The following results about derivatives follow exactly as in the case of reals:(1) Derivative of a constant function is zero andddz(zn) =nzn 1, n N.(2) If , Cthen ( f+ g) = f + g .(3) (Chain Rule)ddzf(g(z)) =f (g(z))g (z) whenever all the terms make much for similarity. To see the difference of COMPLEX derivatives and thederivatives of functions of two real variables we look at the following the functionf:C Rgiven byf(z) =|z| +iythe functionfcan also be thought of as a function point of view the functionfcan also be written asf(x, y) =x2+ thepartial derivatives offare continuous throughoutR2it follows thatfis differentiableeverywhere what happens if we now viewfas a function onCand thinkabout COMPLEX differentiability?

3 It is clear thatfis differentiable at zero aslimh 0|h|2h= lim(h1,h2) (0,0)h21+h22h1+ih2= lim(h1,h2) (0,0)h21+h22h21+h22(h1 ih2) = 0(we have used the important fact that|z|2=zz).On the other handlimh 0|z+h|2 |z|2h= limh 0z h+ zhhdoes not exist forz6= 0aslimh 0hhdoes not LECTURE 2: COMPLEX DIFFERENTIATION AND CAUCHY RIEMANN EQUATIONSSo we need to find a necessary condition for differentiability of a function of acomplex are called CAUCHY - Riemann equations (CR equationforshort) given in the following need the following notation to express the theorem which deals with the realpart and imaginary part of a function of a COMPLEX variable.

4 Letf:C Cbe afunction thenf(z) =f(x, y) =u(x, y) +iv(x, y).The functionsuandvcan be thought of as realvalued functions defined on subsets ofR2and are called real and imaginary part offrespectively (u=Re f, v=Im f).Theorem thatf(z) =f(x+iy) =u(x, y) +iv(x, y)is differentiable atz0=x0+ the partial derivatives ofuandvexist at the pointz0= (x0, y0)andf (z0) =ux(x0, y0) +ivx(x0, y0) =vy(x0, y0) iuy(x0, y0).Thus equating the real and imaginary parts we getux=vy, uy= vx,atz0=x0+iy0( CAUCHY Riemann equations).

5 Differentiable atz0we have by varyinghover the set of real numbersf (z0) =f (x0+iy0) = limh 0u(x0+h, y0) u(x0, y0) +i[v(x0+h, y0) v(x0, y0]h=ux(x0, y0) +ivx(x0, y0),andf (z0) =f (x0+iy0)= limh 0u(x0, y0+h) u(x0, y0) +i[v(x0, y0+h) v(x0, y0)]ih= limh 0v(x0, y0+h) v(x0, y0)h ilimh 0u(x0, y0+h) u(x0, y0)h(as1i= i)=vy(x0, y0) iuy(x0, y0).Now we can just compare the real and imaginary parts off (z0).This completes theproof. Note that the crux of the proof is to approach the point (0,0) through real axis(X-axis) and through imaginary axis (Y-axis) (it is the same way we have shownthat the functionf(z) =zis not differentiable).)

6 Now, CR equations has some magical consequences, some of which is 2: COMPLEX DIFFERENTIATION AND CAUCHY RIEMANN EQUATIONS 3(1) Iff:C Cis such thatf (z) = 0 for allz C,thenfis a constantfunction. This is because, by CR equationux=uy=vx=vy= byMVT of two variable calculusuandvare constant function and hence so isf.(2) Iff:C Cis differentiable everywhere andf(z) is real for allz Cthenfis a constant function. This follows from CR equation asv(x, y) = 0 forallx+iy Cand hence all partial derivatives ofvis also zero and hencethe same is true the functionf(z) =|z|2is not differentiable forz6= CR equations do not give a sufficient criteria for (z) =z2/z,ifz6= 0andf(0) = is easy to see that thisfunction is not differentiable definitionlimh 0f(h) f(0)h= limh 0( h)

7 By choosinghto be real we get the limit to be1and replacinghbyh+ihwesee that the limit is real and imaginary parts offsatisfies CR equations atz= 0(check this!).If we add some more conditions on the partial derivatives ofuandvalong withCR equations then one can conclude that the function is differentiable. We state atheorem (without the proof) for the precise 5.(Converse of CR relations)f=u+ivbe defined onBr(z0)suchthatux, uy, vx, vyexist onBr(z0)and are continuous CRequations thenf (z0)exist andf =ux+ the above result we can immediately check that the functions(1)f(x+iy) =x3 3xy2+i(3x2y y3)(2)f(x+iy) =e ycosx+ie ysinxare differentiable everywhere in the COMPLEX equations can also be expressed in the polar :Usingx=rcos , y=rsin and the chain rule u r= u x x r+ u y y rprovethat the CR equation is equivalent tour=1rv ,vr= 1ru.

8 4 LECTURE 2: COMPLEX DIFFERENTIATION AND CAUCHY RIEMANN EQUATIONSAs in the cartesian case, it can be proved that ifur, u , vr, v are continuous andsatisfies CR equations then the function is differentiable.