[PDF] Topic 3 Notes 3 Line integrals and Cauchys theorem

1 Review of complex numbers

Im z = 1. 2i(z ? ¯z). Thus z is real if and only if ¯z = z and pure imaginary if and only if ¯z = ?z. More importantly we have the following formulas 

Lecture 5. Complex Numbers and Eulers Formula

Figure 1: A complex number z and its conjugate ¯z in complex space. Eg 5.2.1 Given that z1 =3+4i z2 = 1 ? 2i

Math 311 - Spring 2014 Solutions to Assignment # 7 Completion

e?zt dt (Re z > 0). Ans: (a) ?. 1. 2 ? i ln 4;. (b). ?3. 4. + i. 4. ;. (c). 1 z . Solution: (a) We have. ? 2. 1 (1t ? i). 2 dt = ?. 2. 1 ( 1t2 ?. 2i.

Topic 4 Notes 4 Cauchys integral formula

(z ? 2i)2(z + 2i)2 . Let f(z) = 1. (z + 2i)2 . Clearly f(z) is analytic inside C. So by Cauchy's formula for derivatives: ?. C. 1. (z2 + 4)2.

Complex Numbers and Exponentials

(0 1). In this notation

Topic 3 Notes 3 Line integrals and Cauchys theorem

1. 0 t2(1 + i)2(1 + i)dt = 2i(1 + i). 3 . Example 3.2. Compute. ? ? z dz analytic function on an open region A and ? is a curve in A from z0 to z1 then.

Complex Numbers and Exponentials

In this notation the sum and product of two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 is given For example

Complex Variables 05-3 Exam 1 Solutions

http://www.math.pitt.edu/~sparling/053/156053/156053quizzes/156053e1s.pdf

Complex numbers - Exercises with detailed solutions

1. Compute real and imaginary part of z = i ? 4. 2i ? 3 . 2. Compute the absolute value and the conjugate of z = (1 + i)6.

Topic 7 Notes 7 Taylor and Laurent series

The geometric series in this equation has ratio z/w. Therefore the series converges

< 1

6 Residues and Poles - University of California Irvine

z?2 + 1 z?1?2i is also analytic at zero the Laurent series of f(z) has the form f(z) = 3 z + 1 z2 + ? ? n=0 anz n We conclude that f(z) has a pole of order 2 at z0 = 0 and residue Res z=0 f(z) = 3 Similarly f(z) has simple poles (order 1) at z1 = 2 and z2 = 1 +2i with Res z=2 f(z) = 5i Res z=1+2i f(z) = 1 3 e1/z = ? ? n=0 z

Residues and Contour Integration Problems - Texas A&M University

we have that (z+i)g(z) = 1 z i which is analytic and nonzero at z= i Hence g(z) has a simple pole at z= i The residue is thus Res i(g) = lim z! i(z+ i)g(z) = 1 2i = 1 2 i 8 g(z) = ez z3 at z= 0 Ans Res 0(g) = 1 2 Solution Using the power series for ez we see that the Laurent series for g(z) about z= 0 is ez z 3 = 1 + z+ 1 2! z2 + 1

The Complex Plane - Michigan State University

(3) By completing the square z4 4z2 + 4 2i= (z2 2)2 2iand so the equation is equivalent to (z2 2)2 = 2i= (1 + i)2;which is equivalent to z2 2 = 1 + i or z2 2 = 1 i; that is z2 = 3 + i or z2 = 1 i: Hence the solutions are the square roots of 3 + iand 1 i:They are in the form of z= z 1; z= z 2; where z2 1 = 3 + iand z2 2 = 1 ican be found in

Complex Variables Review Problems (Residue Calculus

(z2+1)(z2+2z+3) f(z) has two poles iand 1 + i p 2 in the upper half plane which are both simple Then Res(f(z);i) = 1=8 i=8 and Res(f(z); 1 + i p 2) = 1=8 The Residue Theorem shows that Z 1 1 dx (x2 + 1)(x2 + 2x+ 3) = 2?i( 1=8 i=8 + 1=8) = ? 4 provided we can show that lim R!1 Z C R f(z)dz= 0 where C R is the semicircle jzj= Rwith =z 0 To

How do you calculate I2 C 1 z2dz?

I2 = ? C 1 z2dz. First, in the applet select the function f (z)=1/z^2. Then analize the values of I2 in the following cases: C is any contour from z0 = ? i to z1 = i. What happens when you select Line Segment in the applet? What happens when you select Semicircles?

How to evaluate C1 ZDZ?

Now let’s evaluate ? C1 zdz, where C is the circle x = cost, y = sint, with 0 ? t ? 2?. Figure 2: C: z(t) = cost + isint = eit, with 0 ? t ? 2?. and, z ? (t) = ieit. Thus ? C1 zdz = ?2? 0 (e ? it)ieitdt = i?2? 0 dt = 2?i. Use the following applet to explore numerically the integral Line segments. Semicircles.

What is a contour integral f(z) from a fixed point z0 to Z1?

Although the value of a contour integral of a function f(z) from a fixed point z0 to a fixed point z1 depends, in general, on the path that is taken, there are certain functions whose integrals from z0 to z1 have values that are independent of path, as you have seen in Exercises 2 and 3.

Is z t a real integral of a complex-valued function?

Since C is a contour, z? (t) is also piecewise continuous on a ? t ? b; and so the existence of integral ( 1) is ensured. The right hand side of ( 1) is an ordinary real integral of a complex-valued function; that is, if w(t) = u(t) + iv(t), then in terms of its real and imaginary parts, as well as the differential