Advanced Complex Number Operations

Euler’s Formula

Property A (Euler’s Formula): For any complex number z = a + bi

Proof: For any real number z

The same formula holds for a complex number z (in fact, this can be viewed as the definition of e^z). So, if z = bi, where b is a real number, then

But

i⁰ = 1, i¹ = i, i² = -1, i³ = –i, i⁴ =1, i⁵ = i, etc.

Thus

where we use the infinite sum representations of sin b and cos b for real numbers b.

If z = a + bi, then

Property B (Euler’s identity)

Proof: By Property A

= 1 ⋅ (-1 + i ⋅ 0) = -1

Actually

Log Function

If z is a complex number, by definition, if ln z = w, then z = e^w. Suppose that z = a + bi. As described in Complex Numbers in Polar Format, we can represent z in polar format as

where

Here

Thus

This proves the following property.

Property C

This formula is well defined except for z = 0 since r > 0 otherwise. Note that actually there are infinitely many results, as described below, since

Note that the principal value of the natural log is the one where -π < θ ≤ π.

Examples

Find ln(-1), ln(i) and ln(-i).

The polar form for -1 is r = 1 and θ = π. Thus

ln(-1) = ln 1 + i(π) = iπ

The polar form for i is r = 1 and θ = π/2. Thus

ln(i) = ln 1 + i(π/2) = iπ/2

The polar form for –i is r = 1 and θ = –π/2. Thus

ln(i) = ln 1 + i(-π/2) = –iπ/2

Power of a real number

Property D: For any complex number z = a + bi and any real number c > 0

c^z = c^a(cos(b ln c) + sin(b ln c))

Proof: Using Property A and the fact that z ln c = a ln c + (b + ln c)i

c^z = (e^{ln c})^z = e^{z ln c} = e^{a ln c}(cos(b ln c) + sin(b ln c))

= (e ^{ln c})^a ⋅ (cos(b ln c) + sin(b ln c))

=c^a(cos(b ln c) + sin(b ln c))

It turns out that this property holds even when c < 0. In this case, however, ln c is a complex number.

Multiple Representations

Since

cos(b+2πn) = cos(b)          sin(b+2πn) = sin(b)

for any integer n, from Property A it follows that

e^z = e^a(cos(b+2πn) + sin(b+2πn))

If e^z = w, then w = ln z. Thus, the natural log has infinitely many representations. If z = a + bi is one of these, then so are z = a + (b+2πn)i for any n.

Power function

We now show how to calculate z^w for any complex numbers z = a + bi and w = c + di.

As described above, we can express z in polar format as z = re^θi. Then using Property C

z^w = (e ^{ln z})^w = e^{w ln z} = e^{w (ln r + θi)} = e^{w ln r} ⋅ e^wθi = r^we^wθi

= r^c+di e^(c+di)θi = r^c r^di e^-dθ+cθi = r^c e^{lnr⋅ di} e^-dθ e^cθi = r^ce^-dθ ⋅ e^{(d⋅lnr+cθ)i}

Now by Property A

z^w = r^ce^-dθ ⋅ e^{(d⋅lnr+cθ)i} = r^ce^-dθ (cos φ + i sin φ)

where

φ = d lnr + cθ

Note that all the calculations involve only real numbers.

Examples

Find 2ⁱ, (-2)ⁱ, √i, iⁱ, i^1/i, i^√2

2ⁱ – The polar form for 2 is r = 2 and θ = 0. Thus φ = 1 ln2 + 0 = ln2 ≈ .693147

2ⁱ = 2⁰e^–0 (cos ln2 + i sin ln2) ≈ cos .693147 + i sin .693147 ≈ .769239 + .638961i

(-2)ⁱ – The polar form for -2 is r = 2 and θ = π. Thus φ = 1 ln2 + 0 = ln2 ≈ .693147

(-2)ⁱ = 2⁰e^–1⋅π (cos ln2 + i sin ln2) ≈ e^-π (cos .693147 + i sin .693147)

≈ .043214(.769239 + .638961i) = .033242 + .027612i

√i = i^1/2 – The polar form for i is r = 1 and θ = π/2. Thus φ = 0 + (π/2)(1/2) = π/4

√i = 2⁰e^–0 (cos π/4 + i sin π/4) = √2/2 + i √2/2  ≈ .707107 + .707107i

iⁱ – The polar form for i is r = 1 and θ = π/2. Thus φ = 0 + 0 = 0

iⁱ = 2⁰e^–π/2 (cos 0 + i sin 0) = e^–π/2(1 + 0i) = e^–π/2  ≈  .20788

i^1/i = i^(1/i)(i/i) = i^{i/i^2} = i^i(-1) = i^-i = 1/iⁱ = 1/e^–π/2 = e^π/2 ≈  4.810477

i^√2 – The polar form for i is r = 1 and θ = π/2. Thus φ = 0 + (π/2)√2 ≈ 2.221441

i^√2 ≈ 2⁰e^–0 (cos 2.221441 + i sin 2.221441) ≈ -.6057 + .795693i

nth Root

z^1/n = (re^iθ)^1/n = r^1/n e^iθ/n

As we saw in Complex Numbers in Polar Format, there are actually n solutions

z^1/n = r^1/n e^i(θ+2πk)/n

for k = 0, 1, …, n-1.

For rational exponents m/n where lcd(m,n) = 1

z^m/n = r^m/n e^{i(θ+2πk)m/n}

Thus, once again there are n solutions. For all other cases of z^w there are infinitely many solutions.

Trig Functions

We define the trigonometric functions for complex numbers so that Euler’s formula holds, namely

e^zi = cos z + i sin z

Thus

e^zi + e^–zi = (cos z + i sin z) + (cos (-z) + i sin (-z))

= (cos z + i sin z) + (cos z – i sin z) = 2 cos z

which means that

Similarly

Thus

Also

Finally, note that

If w = arctan z then also w + nπ = arctan z

arctan(z) is not defined for z = nπ/2 for any odd integer n

If w = arcsin z then also w + 2πn = arcsin z

If w = arccos z then also w + 2πn = arccos z

References

Redcrab (2026) Complex numbers online calculator
https://www.redcrab-software.com/en/Calculator/Complex

Proof Wiki (2026) Definition: inverse tangent/complex
https://proofwiki.org/wiki/Definition:Inverse_Tangent/Complex