ComplexLogarithms.pdf - 95 SECTION 3.5 3.5 Complex...

This preview shows page 1 - 3 out of 9 pages.

SECTION 3.5 95 § 3.5 Complex Logarithm Function The real logarithm function ln x is defined as the inverse of the exponential function y = ln x is the unique solution of the equation x = e y . This works because e x is a one-to-one function; if x 1 = x 2 , then e x 1 = e x 2 . This is not the case for e z ; we have seen that e z is 2 πi -periodic so that all complex numbers of the form z + 2 nπi are mapped by w = e z onto the same complex number as z . To define the logarithm function, log z , as the inverse of e z is clearly going to lead to difficulties, and these difficulties are much like those encountered when finding the inverse function of sin x in real-variable calculus. Let us proceed. We call w a logarithm of z , and write w = log z , if z = e w . To find w we let w = u + vi be the Cartesian form for w and z = re θi be the exponential form for z . When we substitute these into z = e w , re θi = e u + vi = e u e vi . According to conditions 1.20, equality of these complex numbers implies that e u = r or u = ln r, and v = θ = arg z. Thus, w = ln r + θi , and a logarithm of a complex number z is log z = ln | z | + (arg z ) i. (3 . 21) We use ln only for logarithms of real numbers; log denotes logarithms of com- plex numbers using base e (and no other base is used). Because equation 3.21 yields logarithms of every nonzero complex number, we have defined the complex logarithm function. It is defined for all z = 0, and because arg z is determined only to a multiple of 2 π , each nonzero complex number has an infinite number of logarithms. For example, log (1 + i ) = ln 2 + ( π/ 4 + 2 ) i = (1 / 2) ln2 + (8 k + 1) πi/ 4 . Thus, to the complex number 1+ i , the logarithm function assigns an infinite number of values, log (1 + i ) = (1 / 2) ln 2 + (8 k + 1) πi/ 4. They all have the same real part, but their imaginary parts differ by multiples of 2 π . In other words, the logarithm function is a multiple-valued function; to each complex number in its domain, it assigns an infinity of values. Example 3.15 Express log (2 - 3 i ) in Cartesian form. Solution Since | 2 - 3 i | = 13, and arg (2 - 3 i ) = 2 - Tan - 1 (3 / 2), log (2 - 3 i ) = ln 13 + [2 - Tan - 1 (3 / 2)] i = 1 2 ln 13 + [2 - Tan - 1 (3 / 2)] i. Some of the properties of the real logarithm function have counterparts in the complex logarithm. For example, if z = x + yi = re θi , then e log z = e ln r + θi = e ln r e θi = re θi = z, (3 . 22a) (as should be expected from the definition of log z ), and
Image of page 1

Subscribe to view the full document.

96 SECTION 3.5 log ( e z ) = log ( e x + yi ) = ln ( e x ) + ( y + 2 ) i ( k an integer) = x + yi + 2 kπi = z + 2 kπi. (3 . 22b) In real analysis the counterpart of this equation is log e x = x . The z + 2 on the right side of 3.22b is a reflection of the facts that log z is multiple-valued and the logarithm is the last operation on the left side of the equation. If z 1 = re θi and z 2 = Re φi , then log ( z 1 z 2 ) = log [ rRe ( θ + φ ) i ] = ln ( rR ) + ( θ + φ + 2 ) i ( p an integer) = (ln r + θi ) + (ln R + φi ) + 2 pπi.
Image of page 2
Image of page 3
  • Winter '17
  • Complex number, Logarithm, branch cuts

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern