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Find the real and imaginary parts u(x,y)and v(x,y)of the following functions.

1z

Short Answer

Expert verified

The real partux,yof the function is xx2+y2and the imaginary partvx,yof the function isyx2+y2 .

Step by step solution

01

Definition of complex number

Complex number is represented by, wherea+ibis the a real number and bis the imaginary number.

02

Solve complex number

Given the function is 1z.

The complex number z can be written as:1x+iy

z=x+iy, where x is a real part and y is an imaginary part.

The function 1zcan be written as

Multiply with complex conjugate of z=x+iyinto denominator and numerator to separate real and imaginary parts.

1z=x-iyx+iyx-iy=x-iyx2-i2y2=x+iyx2+y2

03

Find real and imaginary parts

Simplify the expression future

x+iyx2+y2=xx2+y2+iyx2+y2

Hence the real part is xx2+y2and imaginary part is yx2+y2.

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Most popular questions from this chapter

In equation (7.18), let u (x) be an even function and Ï…(x)be an odd function.

  1. If f(x)=u(x)+iÏ…(x), show that these conditions are equivalent to the equationf*(x)=f(-x) .
  2. Show that

πu(a)=PV∫0∞2xυ(x)x2-a2 dx,      πυ(a)=-PV∫0∞2au(x)x2-a2 dx

These are Kramers-Kroning relations. Hint: To find u(a), write the integral for u(a) in (7.18) as an integral from -∞to 0 plus an integral from 0 to ∞. Then in the to integral -∞to 0, replace x by -x to get an integral from 0 to∞ , and userole="math" localid="1664350095623" υ(-x)=-υ(x) . Add the two to integrals and simplify. Similarly findrole="math" localid="1664350005594" υ(a) .

Question: Verify that the given function is harmonic, and find a functionof which it is the real part. Hint: Use Problem 2.64. For Problem 2, see Chapter 2, Section 17, Problem 19.

ln(1+x)2+y2

Evaluate the integrals by contour integration.

I=∫0xcos(θ)dθ5-4cos(θ)

Use the Cauchy-Riemann conditions to find out whether the functions in Problems 1.1 to 1.21 are analytic.

2z+3z+2

Let f(z) be expanded in the Laurent series that is valid for all z outside some circle, that is,|z|>M(see Section 4). This series is called the Laurent series "about infinity." Show that the result of integrating the Laurent series term by term around a very large circle (of radius > M) in the positive direction, is 2πib1(just as in the original proof of the residue theorem in Section 5). Remember that the integral "around ∞ " is taken in the negative direction, and is equal to 2πi: (residue at ∞). Conclude thatR(∞)=-b1 . Caution: In using this method of computingR(∞) be sure you have the Laurent series that converges for all sufficiently large z.

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