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Last modified on April 30th, 2024

The complex logarithm is an extension of the concept of logarithmic functions involving complex numbers (represented by log z).

Mathematically, written as

log(z) = log(r â‹… e^{iÎ¸}) = ln(r) + i(Î¸ + 2nâ„¼)

Here,

z = r â‹… e^{iÎ¸} = the complex number

r = |z| = the absolute value of z

Î¸ = arg(z) = the argument of z and -â„¼ < Î¸ â‰¤ â„¼

ln(r) = real part

i(Î¸ + 2nâ„¼) = imaginary part

n = multiple branches of the complex logarithm and n Ð„ â„¤

log(z) = log(r â‹… e^{iÎ¸}) = ln(r) + i(Î¸ + 2nâ„¼)

Here,

z = r â‹… e^{iÎ¸} = the complex number

r = |z| = the absolute value of z

Î¸ = arg(z) = the argument of z and -â„¼ < Î¸ â‰¤ â„¼

ln(r) = real part

i(Î¸ + 2nâ„¼) = imaginary part

n = multiple branches of the complex logarithm and n Ð„ â„¤

Thus, it is also written as **log(z) = ln|z| + i(arg (z) + 2nâ„¼)**

Let us consider a non-zero complex number â€˜z,â€™ such that z = e^{w} â€¦..(i)

If Î¸ = arg(z) with -â„¼ < Î¸ â‰¤ â„¼, then â€˜zâ€™ and â€˜wâ€™ are expressed as

z = r â‹… e^{iÎ¸} and w = u + iv, where â€˜uâ€™ is the real part, and â€˜ivâ€™ is the imaginary part

Thus, from equation (i), we get

e^{u + iv} = r â‹… e^{iÎ¸}

â‡’ e^{u} â‹… e^{iv} = r â‹… e^{iÎ¸}

On comparing both sides, we get

e^{u} = r and e^{iv} = e^{iÎ¸}

â‡’ u = ln(r) and v = Î¸ + 2nâ„¼, where n Ð„ â„¤.

It follows that the equation (i) is satisfied if and only if w = ln(r) + i(Î¸ + 2nâ„¼), where n Ð„ â„¤.

Thus, the equation (i) becomes,

z = e^{w}

â‡’ log(z) = w

â‡’ log(z) = ln(r) + i(Î¸ + 2nâ„¼), where n Ð„ â„¤ â€¦..(ii)

â‡’** log(z) = ln|z| + i(arg (z) + 2nÏ€)** â€¦..(iii)

Thus, the logarithm of a non-zero complex number z in its general form is derived,

It is thus an example of a multiple-valued function, where all the multiple values of the complex logarithm have the same real part ln(r) but differ in the imaginary part by â€˜2â„¼’.

**Solve the given complex logarithmic function log(z) for ${z=2\sqrt{3}+2i}$**

Solution:

Here, ${z=2\sqrt{3}+2i}$, then ${r=\sqrt{16}}$ and Î¸ = ${\tan ^{-1}\left( \dfrac{2}{2\sqrt{3}}\right)}$ = ${\dfrac{\pi }{6}}$

Thus, ${\log \left( 2\sqrt{3}+2i\right) =\ln \sqrt{16}+i\left( \dfrac{\pi }{6}+2n\pi \right)}$

â‡’ ${\log \left( 2\sqrt{3}+2i\right) =\ln \sqrt{16}+\left( \dfrac{1}{6}+2n\right) \pi i}$, where n Ð„ â„¤

**Calculate the value of log(1)**

Solution:

As we know, e^{0} = 1

Thus, log(1) = 0

Also, we know e^{2â„¼i} = 1

â‡’ log(1) = 2â„¼i, another possible answer.

Thus, log(1) = 2nâ„¼i, where n Ð„ â„¤

When n = 0, the principal branch (or the principal value) of log(z), represented by Log(z), is obtained from the equation log(z) = ln(r) + i(Î¸ + 2nâ„¼) as:

Log(z) = ln(r) + iÎ¸ â€¦..(iv)

Thus, the equation z = e^{w} can also be written as log(z) = Log(z) + 2nâ„¼i, where n Ð„ â„¤ â€¦..(v)

Here, the principal branch of the log (Log(z)) is evaluated from the principal branch of the arg, where -â„¼ < Î¸ â‰¤ â„¼.

**Solve: Log(-1)**

Solution:

Here, z = -1 = -1 + 0i, then r = 1 and Î¸ = ${\tan ^{-1}\left( \dfrac{0}{-1}\right)}$ = â„¼

Now, Log(-1) = ln(1) + iÎ¸ = 0 + â„¼i = â„¼i

Thus, Log(-1) = â„¼i

For log (z), infinity is the branch point. Since the function log(z) is infinity only at z = 0, the loop must enclose the origin.

Here, the branch cuts along the negative real axis, implying a discontinuity along this axis due to the multivalued nature of log(z).

Let â€˜zâ€™ be a non-zero complex number function such that e^{log(z)} = z â‡’ e^{w log(z)} = z^{w}, where â€˜wâ€™ is another complex number function â€¦..(vi)

Also, log(e^{z}) = z + 2nâ„¼i, where n Ð„ â„¤ â€¦..(vii)

Let us calculate e^{log(z)} and log(e^{z}) for z = 1 + 5i

Here, e^{log(z)} = z = 1 + 5i and log(e^{z}) = z + 2nâ„¼i = 1 + 5i + 2nâ„¼i = 1 + i(5 + 2nâ„¼), for n Ð„ â„¤.

**Find the value of z ^{w}, for the complex numbers z = 1 + i and w = i**

Solution:

As we know, z^{w} = e^{w log(z)}

â‡’ (1 + i)^{i} = e^{i log(1 + i)}

Here, ${r=\sqrt{2}}$ and Î¸ = ${\tan ^{-1}\left( \dfrac{1}{1}\right)}$ = ${\dfrac{\pi }{4}}$Â

log(1 + i) = ${\ln \left( \sqrt{2}\right) +i\left( \dfrac{\pi }{4}+2n\pi \right)}$

Now, e^{i log(1 + i)}

= ${e^{i\left( \ln \left( \sqrt{2}\right) +i\left( \dfrac{\pi }{4}+2n\pi \right) \right) }}$

= ${e^{\left( -\dfrac{\pi }{4}+i\left( \ln \sqrt{2}\right) +2n\pi \right) }}$

= ${e^{-\dfrac{\pi }{4}}\cdot e^{i\ln \sqrt{2}}\cdot e^{2n\pi }}$

= ${e^{-\dfrac{\pi }{4}}\cdot \left( \sqrt{2}\right) ^{i}\cdot e^{2n\pi }}$, for n Ð„ â„¤.

Like the real-valued logarithmic functions, the complex logarithms hold the same properties as discussed below.

Product Rule | log(z_{1} â‹… z_{2}) = log(z_{1}) + log(z_{2}) |

Quotient Rule | ${\log \left( \dfrac{z_{1}}{z_{2}}\right) =\log z_{1}-\log z_{2}}$ |

Power Rule | log(z^{n}) = n â‹… log(z), where n Ð„ â„¤ |

A mapping of a function shows how the elements of the domain are connected with the elements of the range. If w = log(z), the domain is z, and the range is w.

Here, some examples of the mapping of log(z) are:

1) Here, we observe that a point â€˜zâ€™ is mapped to many values of â€˜w,â€™ and the dots are represented as:

log(1) â†’ blue dots, log(8) â†’ red dots, log(i) â†’ blue cross, and log(8i) â†’ red cross.

The values in the principal branch are inside the shaded region in the w-plane, and the values of log(z) for a given â€˜zâ€™ are placed at intervals of â€˜2Ï€iâ€™ in the w-plane.

2) Here, the circles centered on 0 are mapped on the y-axis, and rays from the origin are mapped on the x-axis.

We observe that the circles are mapped to the y-axis in the principal branch, and rays are mapped to the x-axis in the principal region of the w-plane, which is shaded as shown.

Last modified on April 30th, 2024