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X ~ ~o6 C撥d'9I u) ҥ ~ 5 27 ) pR e 9 Associated analyticfunction f (z) Solution Let f (x,y)=u(x,y)iv(x,y)where u and v arereal Thenu =sinx coshy and v =cosx sinhy st ux =cosx coshy =vy, vx =−sinx sinhy =−uy, ie CR conditions hold f (z)= (eix −e−ix)(ey e−y) 4i i (eix e−ix)(ey −e−y) 4 = = eixe−y −e−ixey 2i = ei(xi y)−e−i(xi y) 2i =sinz Problem #2 ( points) Given theimaginarypart,Sinv;1idA = Z 2ˇ 0 Z 1 0 ( ucos2 v usin2 v u4)ududv = Z 2ˇ 0 Z 1 0 (u2 u5)dudv = 2ˇ Z 1 0 (u2 u5)du = 2ˇ u3 3 u6 6 1 0 = ˇ 3

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I also tried to use Liouville's Theorem on $\frac{f(z)}{f(\f Stack Exchange Network Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn,Links with this icon indicate that you are leaving the CDC website The Centers for Disease Control and Prevention (CDC) cannot attest to the accuracy of a nonfederal website Linking to a nonfederal website does not constitute an endorsement by CDC or any of its employees of the sponsors or the information and products presented on the websiteF) dV = 0 Therefore ZZ S F

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Theorem If the 1form f(z)dz is of the form dF, or equivalently the vector eld (uiv;F(z) = f(z 0) then we say fis continuous at z 0 If fis de ned on an open region Athen the phrase'fis continuous on A'means that fis continuous at every point in A As usual, we can rephrase this in terms of functions of (x;y) Fact f(z) = u(x;y) iv(x;y) is continuous i u(x;y) and v(x;y) are continuous as functions of two variables Example 25¹zceqtcr @`d@ jk\jqc <n>;@`f eg@c k@`dgc m bk@ cedge ceokj¥cqq 6xz6!µ= jkdzqgs;@ <n>;z`mnu;eg<

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Where Dis the domain of the parameters uand v, the rectangle 0;1 0;2ˇ Evaluating this integral, we obtain Z Z S FdS = Z Z D hucosv;usinv;u4iuh cosv;Answer Writing out ez as a power series we have f(z) = 1 z z2=2!Imf(z) = v, f(z) = u iv, jf(z)j= p u2 v2 Likewise, if g(z) is another complex function, we can de ne f(z)g(z) and f(z)=g(z) for those zfor which g(z) 6= 0 Some of the most interesting examples come by using the algebraic operations of C For example, a polynomial is an expression of the form P(z) = a nzn a n 1zn 1 a 0;

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S F dS for vector eld F = xi yjz4k and the oriented surface Sthat is the part of the cone z= p x2 y2 beneath the plane z= 1 with downward orientation In other words, nd the ux of F across S Using spherical coordinates, the cone is expressed as x= ˆ p 2 cos ;Z= ˆ p 2 Note that ˚= ˇ=4 We obtain a vector equation r( ;ˆ) = ˆ p 2 cos i ˆ p 2 sin j ˆ p 2 k;R and the power law equation (formula) P = I ×

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C be a holomorphic function such that jfj is the constant function c Check that f is holomorphic Solution If c = 0 ) jfj = f = f = 0, hence f is holomorphic Suppose jfj = c 6= 0 ) f(z) 6= 0 8z 2 C Since f is holomorphic, we have c2 f is holomorphic Now, ff = jfj2 = c2 therefore f = c2 f, which we showed is holomorphic So by159 The Divergence Theorem The Divergence Theorem is the second 3dimensional analogue of Green's Theorem Recall if F is a vector field with continuous derivatives defined on a region D R2 with boundary curve C, then I C F nds = ZZ D rFdA The flux of F across C is equal to the integral of the divergence over its interiorWrite f (z) =u (x,y) iv (x,y) As f (z) is analytic, by CauchyRiemann conditions, u_x = v_y and u_y = v_x, where the subscripts denote partial differentiation By hypothesis, f (z)^2 = u (x,y)^2 v (x,y)^2 = k

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$\begingroup$ @takecare They only need to be equal on a sequence which is convergent in the domain ie the functions must be equal on the sequence and analytic at the limit point The open set version is weaker than needs be $\endgroup$ – N SF(z) = u(x,y)iv(x,y) where z = x iy and u, v are real functions Suppose that f is differentiable at z We can take δz in any direction;F(V) \f • Wf 0g −‰ = f(V) \f!g = f!g This is a contradiction Spring 11, 2 Let hbe a holomorphic function on a connected open set V Prove that if h(z)2 = h(z) for all z2V then his constant on V Find all possible values for such h First of all, V is nonempty1 Let f= h3 2H(V) So for all z2V, f(z) = h(z)2h(z) = h(z)h(z) = jh(z)j2 2R

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4) Explain the MilneThomson's method to construct an analytic function F(z) = u iv when the imaginary part v is given 5) Find an analytic function F(z) = u iv and express it in terms of z if u = x3 3xy2 3x2 3y2 1 6) Find an analytic function F(z) = u iv if, v = eysinx and F(0) = 1>H )p3, ' Dq sq V tR 6 ȅ _ s` , Hs/y z ~ ( wz m !N dσ = 0 The flux along any surface S vanishes as long as 0 is not included in the region surrounded by S (F is not differentiable at 0) C The Divergence Theorem (Sect 168) I The divergence of a vector field in space I The Divergence Theorem in space

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Answer Using the known series for cos(z) we get f(z) = z2 1 1 2!If we write F = ρv, then F is also a vector field on R3 Then, the integral in Equation 7 becomes A surface integral of this form occurs frequently in physics²even when F is not ρv It is called the surface integral (or flux integral) of F over S S ³³Fn dS

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