The Imaginary Number and Transcendental Functions

The imaginary number, i, defined as \(\sqrt{-1}\), is a critical number in mathematics, although it has no real value. Although the number itself has no value, applying some functions or calculations can result in real numbers. For example, the powers of i alternate between real and imaginary numbers:

(1) \(i^1 = i\) and \(i^{4n+1} = i\)

(2) \(i^2 = -1\) and \(i^{4n+2} = -1\)

(3) \(i^3 = -i\) and \(i^{4n+3} = -i\)

(4) \(i^4 = 1\) and \(i^{4n} = 1\) for all integers n > 0

This page contains various calculations with i. The goal is to represent various calculations as a complex number a + bi, where a is a real number and bi is the imaginary number.

First, recall that if a polynomial with integer coefficients has imaginary roots, they always occur as conjugate pairs. Therefore, if a + bi is a root, then abi is a root. This is the only way the imaginary part is eliminated to ensure the coefficients are real. The product of the conjugate pairs is real: \((a + bi)(a - bi) = \) \(a^2 + bi - bi - b^2\cdot i^2 = \) \(a^2 + b^2\).

First, we’ll list some identities and then offer some proofs of some of them.

Exponential Functions

Euler first noted the link between exponential functions and trigonometric functions when he discovered the identity: \(e^{xi} = \cos x + i\cdot\sin x\). We can generalize this for e raised to any complex number:

(1) \(e^{a+bi} = e^a\cdot e^{bi} = e^a(\cos b + i\cdot\sin b)\)

Equation (1) gives the desired result as a + bi. We can also take the logarithm of a complex number:

(2) \(\ln(a+bi) = \) \(\dfrac{1}{2}\ln(a^2+b^2) + \) \( i\cdot \arctan\dfrac{b}{a}\)

If a = 0, then (2) becomes \(\ln(bi) = \ln b + \dfrac{\pi}{2}i\). Here are some other values:

(3) \(i^i = e^{-\pi/2}\)

Note that the imaginary unit raised to the imaginary unit becomes a real number.

(4) \(x^i = e^{(\ln x)i} = \) \( \cos(\ln x) + i\cdot\sin(\ln x)\)

(5) \((a+bi)^i = \) \( e^{-\arctan(b/a)}\times\) \(\left[\cos(\ln\sqrt{a^2+b^2}) + i\cdot\sin(\ln\sqrt{a^2+b^2}) \right]\)

In equation (5), if we let b = 0, we get equation (4). If we let a = 0, and allow arctan(∞) = π/2, then we get \((bi)^i = e^{-\pi/2}\left[\cos(\ln b) + i\cdot\sin(\ln b) \right]\). If we let b = 1, then we get equation (3).

Trigonometric Functions

(6) \(\sin(a + bi) = \) \(\sin a\cdot\cosh b + \) \( i\cdot\cos a\cdot\sinh b\)

(7) \(\cos(a + bi) = \) \(\cos a\cdot\cosh b - \) \( i\cdot\sin a\cdot\sinh b\)

(8) \(\tan(a+bi) = \) \(\dfrac{\tan a + \tan(bi)}{1-\tan a\cdot\tan(bi)} = \) \(\dfrac{\tan a\cdot\text{sech}^2 b+i\cdot\tanh b\cdot\sec^2 a}{1+(\tan a\cdot\tanh b)^2}\)

In equation (8), if we split the denominator between the terms of the numerator, we do get the a + bi form.

(9) \(\arcsin(a+bi) = \) \( -i\ln\left[\sqrt{1-(a+bi)^2}+i(a+bi) \right]\)

Well, equation (9) doesn’t help much since it’s not in the a + bi form. I’ll have to look at this one again. However, this is the equivalence given by Wolfram Alpha.

(10) \(\arcsin a = \dfrac{\pi}{2} - i\cosh^{-1} a\) (for all real a)

(11) \(\arccos(a+bi) = \) \(\dfrac{\pi}{2} + i\ln\left[\sqrt{1-(a+bi)^2}+i(a+bi) \right]\)

(12) \(\arccos a = i\cosh^{-1} a\) (for all real a)

Hyperbolic Functions

(13) \(\sinh(a+bi) = \) \(\sinh a\cdot\cos b + \) \(i\cdot\cosh a\cdot\sin b\)

(14) \(\cosh(a+bi) = \) \(\cosh a\cdot\cos b - \) \( i\cdot\sinh a\cdot\sin b\)

(15) \(\tanh(a+bi) = \) \( \dfrac{\tanh a + \tanh(bi)}{1+\tanh a\cdot\tanh(bi)} = \) \( \dfrac{\tanh a\cdot\text{sec}^2 b+i\cdot\tan b\cdot\text{sech}^2 a}{1+(\tanh a\cdot\tan b)^2}\)

(16) \(\sinh^{-1}(a+bi) = \) \(\ln\left[\sqrt{(a+bi)^2+1}+a+bi \right]\)

(17) \(\cosh^{-1}(a+bi) = \) \(\ln\left[\sqrt{(a+bi)^2-1}+a+bi \right]\)

Algebraic Operations

(18) \(\sqrt{a+bi} = \) \(\pm(a^2+b^2)^{1/4}\times \) \( \left[ \sqrt{\dfrac{1}{2}+\dfrac{a}{2\sqrt{a^2+b^2}}} + i\sqrt{\dfrac{1}{2}-\dfrac{a}{2\sqrt{a^2+b^2}}} \right]\)

(19) \((a+bi)^n = \) \((\sqrt{a^2+b^2})^n\times \) \((\cos(n\theta) + i\cdot\sin(n\theta))\), where θ = arctan(b/a)

(20) \((a+bi)^{1/n} = \) \( (\sqrt{a^2+b^2})^{1/n}\times \) \( \left [\cos\left(\dfrac{\theta+2k\pi}{n}\right) + i\cdot\sin\left(\dfrac{\theta+2k\pi}{n}\right) \right ]\), where θ = arctan(b/a) and k = 0, 1, 2, ..., (n – 1)

Equation (20) gives all the nth roots of a complex number since any given number has n number of roots. Let’s look at some examples.

Example 1

Equations (19) and (20) are unique in that they allow us to find the nth power or nth roots of a complex number. We can solve some problems with these.

Problem: Evaluate (1 + i)1000

First, θ = arctan(1/1) = π/4. The quantity \(\sqrt{a^2+b^2}\) is the radius, r, of the complex number plotted on a polar graph (or the distance from the origin). For our example, \(r=\sqrt{1^2+1^2} = \sqrt{2}\). Putting this together:

(i) \((1+i)^{1000} = \) \((\sqrt{2})^{1000}\times\) \(\left[\cos(1000\pi/4) + i\cdot\sin(1000\pi/4)\right] \)

(ii) \((1+i)^{1000} = \) \(2^{500}\times \) \(\left[\cos(250\pi) + i\cdot\sin(250\pi)\right] \)

The cosine of odd multiples of π is –1 and even multiples is 1. The sine of both odd or even multiples of π is 0. Therefore,

(iii) \((1+i)^{1000} = 2^{500}\cdot(1 + 0) = 2^{500} \)

Example 2

Problem: Evaluate (2+2i)6

Since a and b are the same, we know that θ must be π/4. The radius is \(\sqrt{2^2+2^2} = 2\sqrt{2}\). Putting this together:

(i) \((2+2i)^{6} = \) \(\left(2\sqrt{2}\right)^6\times \) \(\left[\cos(6\pi/4) + i\cdot\sin(6\pi/4)\right] \)

(ii) \((2+2i)^{6} = \) \(512\cdot\left[\cos(3\pi/2) + i\cdot\sin(3\pi/2)\right] \)

Cosine of 3π/2 = 0 and sine of 3π/2 = –1. Therefore:

(iii) \((2+2i)^{6} = 512(0 - i) = -512i\)

If the complex number is of the form a + ai and it is raised to an even power, we can easily do this calculation. We first raise it to the second power, which eliminates the real part.

(i) \((a + ai)^{2k} =\) \([(a+ai)^2]^k\) (for k is any integer > 0)

(ii) \([a^2 + 2a^{2}i + (ai)^2]^k =\) \((2a^{2}i)^k\)

We can write (ii) as \((\sqrt{2}a)^{n}i^{n/2}\), where n is any even number. In example 2, our a = 2 and n = 6. So, \((2\sqrt{2})^{6}i^{3} = -512i\).

Example 3

Problem: Evaluate \(\left(1-\sqrt{3}i\right)^{2013}\)

The angle measure is \(\theta = \arctan\dfrac{-\sqrt{3}}{1} = -\dfrac{\pi}{3}\). In polar form, our complex number lies in the 4th quadrant. The distance from the origin is \(r = \sqrt{3 + 1} = 2\). Putting it together.

(i) \(\left(1-\sqrt{3}i\right)^{2013} = \) \(2^{2013}\times\) \(\left[\cos(-2013\pi/3) + i\cdot\sin(-2013\pi/3)\right]\)

(ii) \(\left(1-\sqrt{3}i\right)^{2013} = \) \(2^{2013}\times\) \(\left[\cos(-671\pi) + i\cdot\sin(-671\pi)\right]\)

Cosine of –671π = –1 because 671 is an odd multiple of π and sine of –671π = 0 because 671 is an odd multiple of π. Therefore:

(iii) \(\left(1-\sqrt{3}i\right)^{2013} = 2^{2013}(-1 + 0) = -2^{2013}\)

Proofs

Proof of equation (2). First, we convert a + bi to its trigonometric form: \(a+bi = \sqrt{a^2+b^2}(\cos\theta + i\cdot\sin\theta)\), where θ = arctan(b/a). Now, we apply the log to both equivalents:

(i) \(\ln(a+bi) = \) \(\ln\left[ \sqrt{a^2+b^2}(\cos\theta + i\cdot\sin\theta) \right]\)

(ii) \(\ln(a+bi) = \) \(\dfrac{1}{2}\ln(a^2+b^2) + \) \(\ln(\cos\theta + i\cdot\sin\theta)\)

From equation (1), we can rewrite \(\cos\theta + i\cdot\sin\theta\) as \(e^{\theta i}\). Therefore:

(iii) \(\ln(a+bi) = \) \(\dfrac{1}{2}\ln(a^2+b^2) + \) \(\ln(e^{\theta i}) =\) \(\dfrac{1}{2}\ln(a^2+b^2) + \theta i\)

We already defined θ = arctan(b/a). Substitution gives the final result.

(iv) \(\ln(a+bi) = \) \(\dfrac{1}{2}\ln(a^2+b^2) + \) \(i\cdot\arctan\dfrac{b}{a}\)