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Complex Numbers5 min read

Logarithm of Complex Numbers for JEE Main – Essential Guide

Complex logarithms appear occasionally in JEE Main, often yielding surprising results like $i^i$ being real. Mastering these formulas provides quick solutions to otherwise tricky problems.

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Logarithm of Complex Numbers for JEE Main – Essential Guide

Introduction

Complex logarithms appear occasionally in JEE Main, often yielding surprising results like iii^i being real. Mastering these formulas provides quick solutions to otherwise tricky problems.


The Core Formula (Memorize This!)

For z=x+iy0z = x + iy \neq 0:

log(z)=lnz+iarg(z)\boxed{\log(z) = \ln|z| + i\arg(z)}

Where:

  • z=x2+y2|z| = \sqrt{x^2 + y^2} (modulus)
  • arg(z)\arg(z) = angle in radians (principal value: π<θπ-\pi < \theta \leq \pi)

Principal Value: The value when arg(z)\arg(z) is taken in (π,π](-\pi, \pi].


Essential Examples (JEE Favorites)

1. log(i)\log(i)

i=eiπ/2i = e^{i\pi/2}log(i)=0+iπ2=iπ2\log(i) = 0 + i\cdot\frac{\pi}{2} = \frac{i\pi}{2}

2. log(1)\log(-1)

1=eiπ-1 = e^{i\pi}log(1)=0+iπ=iπ\log(-1) = 0 + i\pi = i\pi

3. log(1+i)\log(1+i)

1+i=2eiπ/41+i = \sqrt{2}e^{i\pi/4}log(1+i)=12ln2+iπ4\log(1+i) = \frac{1}{2}\ln 2 + \frac{i\pi}{4}


The iii^i Surprise – A Real Number!

ii=eilog(i)=eiiπ2=eπ/20.2079i^i = e^{i\log(i)} = e^{i\cdot\frac{i\pi}{2}} = e^{-\pi/2} \approx 0.2079

Key Insight: iii^i is purely real despite having imaginary base and exponent!


General Power Formula

For complex z,wz, w: zw=ewlog(z)\boxed{z^w = e^{w\log(z)}}

Step-by-step:

  1. Find log(z)=lnz+iarg(z)\log(z) = \ln|z| + i\arg(z)
  2. Multiply by ww: wlog(z)w\log(z)
  3. Compute e(result)e^{\text{(result)}} using Euler's formula if needed

JEE PYQs & Patterns Solved

PYQ 1: JEE Main 2019 Pattern

Find principal value of iii^i

30-second solution: log(i)=iπ/2\log(i) = i\pi/2 ii=eiiπ/2=eπ/2i^i = e^{i\cdot i\pi/2} = e^{-\pi/2}

Answer: eπ/2e^{-\pi/2}


PYQ 2: JEE Advanced Pattern

Find (1)i(-1)^i

Solution: log(1)=iπ\log(-1) = i\pi (1)i=eiiπ=eπ(-1)^i = e^{i\cdot i\pi} = e^{-\pi}

Answer: eπe^{-\pi} (also real!)


PYQ 3: Value Determination

If log(x+iy)=a+ib\log(x+iy) = a+ib, find a,ba, b

Solution: log(x+iy)=lnx2+y2+itan1(y/x)\log(x+iy) = \ln\sqrt{x^2+y^2} + i\tan^{-1}(y/x) So a=12ln(x2+y2)a = \frac{1}{2}\ln(x^2+y^2), b=tan1(y/x)b = \tan^{-1}(y/x) (with quadrant adjustment)


Quick Reference Table

zzlog(z)\log(z) (Principal Value)
1100
1-1iπi\pi
iiiπ/2i\pi/2
i-iiπ/2-i\pi/2
1+i1+i12ln2+iπ/4\frac{1}{2}\ln 2 + i\pi/4
3+i\sqrt{3}+iln2+iπ/6\ln 2 + i\pi/6
22ln2\ln 2

Important Properties (with Caveats)

1. Product Rule (modulo 2πi2\pi i)

log(z1z2)=log(z1)+log(z2)+2πik\log(z_1z_2) = \log(z_1) + \log(z_2) + 2\pi i k

2. Power Rule (for integer nn)

log(zn)=nlog(z)+2πik\log(z^n) = n\log(z) + 2\pi i k

3. Conjugate Property

log(zˉ)=log(z) if arg(z)π\log(\bar{z}) = \overline{\log(z)} \ \text{if } \arg(z) \neq \pi


Special Results Worth Memorizing

  1. ii=eπ/2i^i = e^{-\pi/2} ≈ 0.2079 (real)
  2. (1)i=eπ(-1)^i = e^{-\pi} ≈ 0.0432 (real)
  3. 2i=cos(ln2)+isin(ln2)2^i = \cos(\ln 2) + i\sin(\ln 2) (on unit circle)
  4. eiπ=1e^{i\pi} = -1 (Euler's identity)

Problem Types & Strategies

Type 1: Direct Evaluation

Example: Find log(1i3)\log(1-i\sqrt{3}) Strategy: Convert to polar form reiθre^{i\theta}, then log(z)=lnr+iθ\log(z) = \ln r + i\theta

Solution: 1i3=2eiπ/31-i\sqrt{3} = 2e^{-i\pi/3} log(1i3)=ln2iπ/3\log(1-i\sqrt{3}) = \ln 2 - i\pi/3


Type 2: Complex Powers

Example: Find (1+i)1i(1+i)^{1-i} Strategy: Use zw=ewlog(z)z^w = e^{w\log(z)}

Solution: log(1+i)=12ln2+iπ/4\log(1+i) = \frac{1}{2}\ln 2 + i\pi/4 (1i)log(1+i)=(expand)(1-i)\log(1+i) = \text{(expand)} (1+i)1i=e(result)(1+i)^{1-i} = e^{\text{(result)}}


Type 3: Equation Solving

Example: Solve ez=1+ie^z = 1+i Strategy: Take log: z=log(1+i)+2πikz = \log(1+i) + 2\pi i k

Solution: z=12ln2+i(π/4+2πk)z = \frac{1}{2}\ln 2 + i(\pi/4 + 2\pi k)


Common Mistakes to Avoid

  1. Forgetting 2πik2\pi i k: In general solutions, include kZk \in \mathbb{Z}
  2. Wrong quadrant: arg(x+iy)=tan1(y/x)\arg(x+iy) = \tan^{-1}(y/x) needs quadrant adjustment
  3. Assuming real log rules: log(z1z2)log(z1)+log(z2)\log(z_1z_2) \neq \log(z_1)+\log(z_2) exactly
  4. Principal value range: Remember (π,π](-\pi, \pi]

Practice Problems

Problem 1: Find log(i)\log(-i)

Solution: i=eiπ/2-i = e^{-i\pi/2} log(i)=0iπ/2=iπ/2\log(-i) = 0 - i\pi/2 = -i\pi/2


Problem 2: Find i2ii^{2i}

Solution: i2i=e2ilog(i)=e2iiπ/2=eπi^{2i} = e^{2i\log(i)} = e^{2i\cdot i\pi/2} = e^{-\pi}


Problem 3: Solve ez=2e^z = -2

Solution: z=log(2)=ln2+i(π+2πk)z = \log(-2) = \ln 2 + i(\pi + 2\pi k)


Advanced Insight: Multi-valued Nature

The complete logarithm is: log(z)=lnz+i(arg(z)+2πk), kZ\log(z) = \ln|z| + i(\arg(z) + 2\pi k), \ k \in \mathbb{Z}

JEE typically uses principal value (k=0k=0), but be aware of the general form.


Exam Strategy

  1. Recognize pattern: log(i),log(1),ii\log(i), \log(-1), i^i appear frequently
  2. Use polar form: z=reiθz = re^{i\theta} simplifies everything
  3. For zwz^w: Always use ewlog(z)e^{w\log(z)}
  4. Check domain: z0z \neq 0 for log(z)\log(z)
  5. Time: 1-2 minutes max for these problems

Final Tip: Remember ii=eπ/2i^i = e^{-\pi/2} as a surprising real number — a favorite JEE fact!


Related: Complex Powers | Euler's Formula

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Practice Complex Numbers PYQs