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). 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 (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? 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).

3 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. 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).

4 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 ). 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).)

5 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).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.

6 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) 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.

7 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 .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.