Complex Logarithmic Functions

Principle Natural Logarithmic Function 

Let 𝑧 = π‘₯ + 𝑦𝑖 be a complex number then the principle natural complex logarithmic function is defined as                                    𝐿𝑛 𝑧 = 𝑙𝑛|𝑧| + π‘–π΄π‘Ÿπ‘” 𝑧 (𝑧 ≠ 0) 

This is also called complex logarithmic formula. 

General Natural Logarithmic Function 

Let 𝑧 = π‘₯ + 𝑦𝑖 be a complex number then the general natural complex logarithmic function is defined as 

                                 π‘™π‘› 𝑧 = 𝐿𝑛|𝑧| ± 2π‘›πœ‹π‘– (𝑛 = 1,2,3, … ) 

i.e.                            𝑙𝑛 𝑧 = 𝑙𝑛|𝑧| + π‘–π΄π‘Ÿπ‘” 𝑧 ± 2π‘›πœ‹π‘– (𝑧 ≠ 0, 𝑛 = 1,2,3, … ) 

Note:

If z is positive real, then Arg z = 0, and Ln z becomes identical with the real natural logarithm known from calculus. If z is negative real (so that the natural logarithm of calculus is not defined!), then Arg z = `\pi` and
        𝐿𝑛 𝑧 = 𝑙𝑛|𝑧| + `\pi`i                                (z negative real)
In a simple logarithm function, we can't take the natural logarithm of negative numbers but in a complex logarithmic function, we can take natural log of negative numbers.


For example 1: 

𝐿𝑛(2) = 0.6931                                      (z is positive real) 
𝐿𝑛(−2) = 𝑙𝑛|−2| + πœ‹π‘– = 0.6931 + πœ‹π‘–       (z is negative real) 

Example 2: 

Find all values of ln(3 − 4𝑖) and show some of these values graphically.
Solution: Let 𝑧 = 3 − 4i
`|z| = \sqrt(3^2 + (-4)^2) = 5`
`\theta = - \tan^-1 |\frac{-4}{3}| = 0.9273`
ln 𝑧 = 𝑙𝑛|𝑧| + π‘–πœƒ = 𝑙𝑛5 + 𝑖(−0.9273) 
Therefore all values of ln 𝑧 are: 
ln 𝑧 = 1.609438 − 0.9273𝑖 ± 2π‘›πœ‹π‘– (𝑛 = 0,1,2,3, … . ) 
The principle value of ln 𝑧 is: 
𝐿𝑛 𝑧 = 1.609438 − 0.9273i
Logarithmic graph

Example 3: 

Find all values of ln(4 + 3𝑖) and show some of these values graphically. 
Solution: Let 𝑧 = 4 + 3i
`|z| = \sqrt(4^2 + 3^2) = 5`
`\theta = - \tan^-1 |\frac{3}{4}| = 0.6435`
ln 𝑧 = 𝑙𝑛|𝑧| + π‘–πœƒ = 𝑙𝑛5 + 𝑖(0.6435) 
Therefore all values of ln 𝑧 are: 
ln 𝑧 = 1.609438 + 0.6435𝑖 ± 2π‘›πœ‹π‘– (𝑛 = 0,1,2,3, … . ) 
The principle value of ln 𝑧 is: 
𝐿𝑛 𝑧 = 1.609438 + 0.6435i
graph

Example 4: 

Find all values of ln(`e^i`) and show some of these values graphically.
Solution: Let 𝑧 = ln(`e^i`)
= ln(π‘π‘œπ‘ 1 + 𝑖𝑠𝑖𝑛1) = ln (0.5403 + 0.8415𝑖)
`|z| = \sqrt(\cos^2 1 + \sin^2 1) = 1`
`\theta = - \tan^-1 |\frac{\sin1}{\cos1}| = 1`
ln 𝑧 = 𝑙𝑛|𝑧| + π‘–πœƒ = 𝑙𝑛1 + 𝑖(1) 
Therefore all values of ln 𝑧 are: 
ln 𝑧 = 0 + 𝑖 ± 2π‘›πœ‹π‘– (𝑛 = 0,1,2,3, … . ) 
The principle value of ln 𝑧 is: 
𝐿𝑛 𝑧 = `i`
graph

Example 5: 

Solve for ‘z’ if ln 𝑧 = `-\frac{\pi}{2}i`
Solution:  Given 𝑙𝑛𝑧 = `-\frac{\pi}{2}i`
`e^lnz = e^(-\frac{\pi}{2}i)`
z = `e^(-\frac{\pi}{2}i)` = `\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2}) = 0 - 1i`
`z = -i`


Example 6: 

Solve for ‘z’ if ln 𝑧 = 0.6 + 0.4𝑖 
Solution:   Given 𝑙𝑛𝑧 = 0.6 + 0.4i
`e^lnz = e^(0.6 + 0.4i)`
`implies     z = e^(0.6 + 0.4i)`
z = `e^0.6 . e^0.5i = e^0.6` (cos(0.4) + 𝑖𝑠𝑖𝑛(0.4)) = `1.6783 + 0.7096i`