If we have to find the square root of negative numbers, then complex numbers are useful. The sum of a real number and an imaginary number is referred to as complex numbers.

Complex number = real number + imaginary number.

For instance, 5 + 2i

In this, 5 is the real number

Whereas, 2i is the imaginary part

Representation of a complex number:-

z = a+ ib,

Where, 

z represents the complex number

a represents the real part

b is the imaginary part

ib is the imaginary number 

Also, if we want to find the negative roots of the quadratic equation, then complex numbers are very useful.

The logarithm of complex number

Let z and w are two complex numbers,

Connected by z= ew 

                       ew= z,

Then, we can say that w is a logarithm of z with base

w = logez

Note: When no base is mentioned, base e is always understood.

We know that,

eiθ = cosθ + isinθ

1) θ = π

   eiπ = cosπ + isinπ      

        = -1 + 0                

        = -1

(Note- cosnπ = (-1n)

           sinnπ = 0)

2) θ = 2nπ

   ei*2nπ = cos2nπ + isin2nπ

           = 1+0

           = 1

Also, 

ax = N, then x is the logarithm of N with a base ‘a’,

Written as x = logaN

2x = 10, then x = log210

Moreover, we know, 

Z = x + iy

y is a real number

i is an imaginary number

W = -1+ √1i / 2

Now, to show that logez is a many-valued fn,

z = ew = w = logez is the definition of the logarithm,

Let n€z

2nπi = cos2nπ + isin2nπ

      = 1 + 0i

      = 1

Moreover, e2nπi = 1           -1st equation

z = ew * 1

   = ew * e2nπi                                – from equation 1st

   = ew+2nπi

w + 2nπi = logez               f n € z

Logez = w + 2nπi               n € z

General and principal value of logez

If z = ew,  then the value w + 2nπi is called the general value of logez

I.e. Logez = w + 2nπi

Logez = logez  + 2nπi

Now, Put n = 0

Logez = logez      – principal value

How does it works?

• First, we find principal value and add 2nπi = General value

logez + 2nπi = Logez

Some properties of logarithm

• logz1 * z2  = logz1 + logz2
• logz1 / logz2 = logz1 – logz2
• log (z1)z2 = z2logz1
• logz2z1 = logz1 / logz2
• loge (x + iy) = 1/2log (x² + y²) + i tan-1 y / x
• loge (x – iy) = 1/2log (x² + y²) + i tan-1 (-y / x)

Examples of logarithm of complex number

1. Q) Find the general and principal value of the logarithm of 

1. 1+i√3
2. -5

1)1+i√3 = r (cosθ + sinθ)

Here, r = √12 + (√32) = 2

         θ = tan-1 √3 = π/3

Therefore, log (1 + i√3)

= log2 + (2nπ + π/3 )            – General

Therefore, log (1 + i√3)

= log2 + iπ/3                        – particular

2) -5 = x + iy

  r (cosθ + sinθ)

  r (cosπ + sinπ)

  5 (cosπ + sinπ)

Therefore, r = 5, θ = π

Log (-5) = log5 + i (2nπ + π)  – General

Log (-5) = log5 + iπ               –  particular

1. Q) Show that loge ( a + bi / a – bi ) = 2i tan-1 b/a

loge ( a + ib)  – loge ( a – ib)

{1/2log (a² + b² ) + itan-1 (b/a)}

-{1/2log (a² + b² ) + itan-1 (- b/a)}

1/2log (a² + b² ) + itan-1 (b/a) – 1/2log (a² + b² ) + itan-1 b/a

= 2 itan-1 b/a

Conclusion

In this topic, we have discussed the logarithm of complex numbers, their definition, showed that logez is a many-valued fn, discussed the General and principal value of logez, and showed how it works. Moreover, mentioned some important properties of the logarithm and rules of the logarithm.

 Also, we have discussed various examples of the logarithm of complex numbers step by step for the convenience of students. So, students can revise their concepts related to the logarithm of Complex Numbers from here before appearing for the exam.