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MTH 362: Advanced Engineering Lecture 3

Jonathan A. Ch´avez Casillas1

1University of Rhode Island Department of Mathematics

September 13, 2017

Jonathan Ch´avez Logarithm Function

1 Logarithm Function

Jonathan Ch´avez Logarithm Function Logarithm Function

The of z = x + iy is denoted by ln z (sometimes also by log z) and is defined as the inverse of the . That is, w = ln z is defined for z , 0 by the relation

ew = z

Assume, in the previous relation that w = u + iv and z = reiθ. Then, substituting back,

eu+iv = reiθ

and thus, since eu+iv = eu · eiv , we get that

eu = r and eiv = eiθ

which imply that u = ln r and v = θ

Jonathan Ch´avez Logarithm Function Logarithm Function

In conclusion, if z = reiθ, and we want to compute w = ln z,

w = ln z = ln r + iθ,

where ln r is the real natural logarithm of the r.

The main problem with the complex logarithm is that the argument is defined up to multiples of 2π. Then, the complex natural logarithm ln z, is infinitely many-valued. However, for z = reiθ, we can define the principal logarithm as Ln(z) = ln r + iArg(z), where Arg(z) is the principal argument of z. In that sense

ln(z) = Ln(z) ± 2nπn = 0, 1, 2,...

Jonathan Ch´avez Logarithm Function Complex Logarithm, continuation

Two important cases need to be mention.

If Im(z) = 0 and Re(z) > 0, that is, z is on the positive real axis, then Ln(z) reduces to the normal real logarithm.

If Im(z) = 0 and Re(z) < 0, that is, z is on the negative real axis, the real logarithm is not defined. Then Arg(z) = π (why is not −π?) and Ln(z) = ln(|z|) + iπ Another important relationship is that (prove it!)

eLn(z) = z but eln(z) = z ± 2nπi, n = 0, 1,...

Jonathan Ch´avez Logarithm Function Properties of the Complex Logarithm

Exercises: Show that

ln(z1z2) = ln(z1) + ln(z2)

ln(z1/z2) = ln(z1) − ln(z2)

Home Exercise: Show that the above relations hold in the sense that “The value obtained on the left hand side is contained in the possible set of values of the right, but may not be true for a unique chosen value on the right hand side”. Hint: Use z1 = z2 = −1 and choose the principal logarithm on both sides.

Jonathan Ch´avez Logarithm Function General Powers

Exercises: The general power of a z is defined by

zc = ec ln(z)

Since ln(z) is infinitely valued, then so does zc . However, we can restrict to the of zc to be defined as ecLn(z). There are three important:

If c = n = 1, 2,..., then zn is single valued and coincides with the real . If c = n = −1, −2,..., then the situation is the same as above.

If c = 1/n, n = 2, 3,..., then √ zc = n z = e1/n ln(z) And the exponent is determined upon multiples of 2π/n, which will give us the n− roots as seen before.

Jonathan Ch´avez