# Imaginary unit

i in the complex or cartesian plane. Real numbers lie on the horizontal axis, and imaginary numbers lie on the vertical axis

The imaginary unit or unit imaginary number, denoted as i, is a mathematical concept which extends the real number system to the complex number system , which in turn provides at least one root for every polynomial P(x) (see algebraic closure and fundamental theorem of algebra). The imaginary unit's core property is that i2 = −1. The term "imaginary" is used because there is no real number having a negative square.

There are in fact two complex square roots of −1, namely i and i, just as there are two complex square roots of every other real number, except zero, which has one double square root.

In contexts where i is ambiguous or problematic, j or the Greek ι (see alternative notations) is sometimes used. In the disciplines of electrical engineering and control systems engineering, the imaginary unit is often denoted by j instead of i, because i is commonly used to denote electric current in these disciplines.

For a history of the imaginary unit, see Complex number: History.

## Definition

The powers of i return cyclic values:
 $\ldots$ (repeats the pattern from blue area) $i^{-3} = i\,$ $i^{-2} = -1\,$ $i^{-1} = -i\,$ $i^0 = 1\,$ $i^1 = i\,$ $i^2 = -1\,$ $i^3 = -i\,$ $i^4 = 1\,$ $i^5 = i\,$ $i^6 = -1\,$ $\ldots$ (repeats the pattern from blue area)

The imaginary number i is defined solely by the property that its square is −1:

$i^2 = -1 \ .$

With i defined this way, it follows directly from algebra that i and i are both square roots of −1.

Although the construction is called "imaginary", and although the concept of an imaginary number may be intuitively more difficult to grasp than that of a real number, the construction is perfectly valid from a mathematical standpoint. Real number operations can be extended to imaginary and complex numbers by treating i as an unknown quantity while manipulating an expression, and then using the definition to replace any occurrence of i 2 with −1. Higher integral powers of i can also be replaced with i, 1, i, or −1:

$i^3 = i^2 i = (-1) i = -i \,$
$i^4 = i^3 i = (-i) i = -(i^2) = -(-1) = 1 \,$
$i^5 = i^4 i = (1) i = i \,$

Similarly, as with any non-zero real number:

$i^0 = i^{1-1} = \frac{i}{i} = 1 \,$
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## i and −i

Being a quadratic polynomial with no multiple root, the defining equation x2 = −1 has two distinct solutions, which are equally valid and which happen to be additive and multiplicative inverses of each other. More precisely, once a solution i of the equation has been fixed, the value i, which is distinct from i, is also a solution. Since the equation is the only definition of i, it appears that the definition is ambiguous (more precisely, not well-defined). However, no ambiguity results as long as one of the solutions is chosen and fixed as the "positive i". This is because, although i and i are not quantitatively equivalent (they are negatives of each other), there is no algebraic difference between i and i. Both imaginary numbers have equal claim to being the number whose square is −1. If all mathematical textbooks and published literature referring to imaginary or complex numbers were rewritten with i replacing every occurrence of +i (and therefore every occurrence of i replaced by −(−i) = +i), all facts and theorems would continue to be equivalently valid. The distinction between the two roots x of x2 + 1 = 0 with one of them as "positive" is purely a notational relic; neither root can be said to be more primary or fundamental than the other.

The issue can be a subtle one. The most precise explanation is to say that although the complex field, defined as [x]/(x2 + 1), (see complex number) is unique up to isomorphism, it is not unique up to a unique isomorphism — there are exactly 2 field automorphisms of [x]/(x2 + 1) which keep each real number fixed: the identity and the automorphism sending x to x. See also Complex conjugate and Galois group.

A similar issue arises if the complex numbers are interpreted as 2 × 2 real matrices (see matrix representation of complex numbers), because then both

$X = \begin{pmatrix} 0 & -1 \\ 1 & \;\;0 \end{pmatrix}$     and     $X = \begin{pmatrix} \;\;0 & 1 \\ -1 & 0 \end{pmatrix}$

are solutions to the matrix equation

$X^2 = -I = - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & \;\;0 \\ \;\;0 & -1 \end{pmatrix}. \$

In this case, the ambiguity results from the geometric choice of which "direction" around the unit circle is "positive" rotation. A more precise explanation is to say that the automorphism group of the special orthogonal group SO (2, ) has exactly 2 elements — the identity and the automorphism which exchanges "CW" (clockwise) and "CCW" (counter-clockwise) rotations. See orthogonal group.

All these ambiguities can be solved by adopting a more rigorous definition of complex number, and explicitly choosing one of the solutions to the equation to be the imaginary unit. For example, the ordered pair (0, 1), in the usual construction of the complex numbers with two-dimensional vectors.

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## Proper use

The imaginary unit is sometimes written −1 in advanced mathematics contexts (as well as in less advanced popular texts). However, great care needs to be taken when manipulating formulas involving radicals. The notation is reserved either for the principal square root function, which is only defined for real x ≥ 0, or for the principal branch of the complex square root function. Attempting to apply the calculation rules of the principal (real) square root function to manipulate the principal branch of the complex square root function will produce false results:

$-1 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{(-1) \cdot (-1)} = \sqrt{1} = 1$    (incorrect).

Attempting to correct the calculation by specifying both the positive and negative roots only produces ambiguous results:

$-1 = i \cdot i = \pm \sqrt{-1} \cdot \pm \sqrt{-1} = \pm \sqrt{(-1) \cdot (-1)} = \pm \sqrt{1} = \pm 1$   (ambiguous).

Similarly:

$\frac{1}{i} = \frac{\sqrt{1}}{\sqrt{-1}} = \sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}} = \sqrt{-1} = i$    (incorrect).

The calculation rules

$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$

and

$\frac{\sqrt{a}} {\sqrt{b}} = \sqrt{\frac{a}{b}}$

are only valid for real, non-negative values of a and b.

These problems are avoided by writing and manipulating i7, rather than expressions like −7. For a more thorough discussion, see Square root and Branch point.

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## Properties

### Square roots

The two square roots of i in the complex plane

The square root of i can be expressed as either of two complex numbers[nb 1]

$\pm \sqrt{i} = \pm \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \right) = \pm \frac{\sqrt{2}}2 (1 + i).$

Indeed, squaring the right-hand side gives

\begin{align} \left( \pm \frac{\sqrt{2}}2 (1 + i) \right)^2 \ & = \left( \pm \frac{\sqrt{2}}2 \right)^2 (1 + i)^2 \ \\ & = \frac{1}{2} (1 + 2i + i^2) \\ & = \frac{1}{2} (1 + 2i - 1) \ \\ & = i. \ \\ \end{align}

This result can also be derived with Euler's formula

$e^{ix} = \cos(x) + i\sin(x) \,$

by substituting x = π/2, giving

$e^{i(\pi/2)} = \cos(\pi/2) + i\sin(\pi/2) = 0 + i1 = i\,\! .$

Taking the square root of both sides gives

$\pm \sqrt{i} = e^{i(\pi/4)} \,\! ,$

which, through application of Euler's formula to x = π/4, gives

\begin{align} \pm \sqrt{i} & = \cos(\pi/4) + i\sin(\pi/4) \\ & = \frac{1}{\pm \sqrt{2}} + \frac{i}{\pm \sqrt{2}}\\ & = \frac{1+i}{\pm \sqrt{2}}\\ & = \pm \frac{\sqrt{2}}2 (1 + i).\\ \end{align}

Similarly, the square root of i can be expressed as either of two complex numbers using Euler's formula:

$e^{ix} = \cos(x) + i\sin(x) \,$

by substituting x = 3π/2, giving

$e^{i(3\pi/2)} = \cos(3\pi/2) + i\sin(3\pi/2) = 0 - i1 = -i\,\! .$

Taking the square root of both sides gives

$\pm \sqrt{-i} = e^{i(3\pi/4)} \,\! ,$

which, through application of Euler's formula to x = 3π/4, gives

\begin{align} \pm \sqrt{-i} & = \cos(3\pi/4) + i\sin(3\pi/4) \\ & = -\frac{1}{\pm \sqrt{2}} + i\frac{1}{\pm \sqrt{2}}\\ & = \frac{-1 + i}{\pm \sqrt{2}}\\ & = \pm \frac{\sqrt{2}}2 (i - 1).\\ \end{align}

Multiplying the square root of i by i also gives:

\begin{align} \pm \sqrt{-i} = (i)\cdot (\pm\frac{1}{\sqrt{2}}(1 + i)) \\ & = \pm\frac{1}{\sqrt{2}}(1i + i^{2})\\ & = \pm\frac{\sqrt{2}}{2}(i - 1)\\ \end{align}

### Multiplication and division

Multiplying a complex number by i gives:

$i\,(a + bi) = ai + bi^2 = -b + ai.$

(This is equivalent to a 90° counter-clockwise rotation of a vector about the origin in the complex plane.)

Dividing by i is equivalent to multiplying by the reciprocal of i:

$\frac{1}{i} = \frac{1}{i} \cdot \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i.$

Using this identity to generalize division by i to all complex numbers gives:

$\frac{a + bi}{i} = -i\,(a + bi) = -ai - bi^2 = b - ai.$

(This is equivalent to a 90° clockwise rotation of a vector about the origin in the complex plane.)

### Powers

The powers of i repeat in a cycle expressible with the following pattern, where n is any integer:

$i^{4n} = 1\,$
$i^{4n+1} = i\,$
$i^{4n+2} = -1\,$
$i^{4n+3} = -i.\,$

This leads to the conclusion that

$i^n = i^{n \bmod 4}\,$

where mod represents the modulo operation.  Equivalently:

$i^n = \cos(n\pi/2)+i\sin(n\pi/2)$

#### i raised to the power of i

Making use of Euler's formula, ii is

$i^i = \left( e^{i (2k \pi + \pi/2)} \right)^i = e^{i^2 (2k \pi + \pi/2)} = e^{- (2k \pi + \pi/2)}$ where $k \in \mathbb{Z}$, the set of integers.

The principal value (for k = 0) is e−π/2 or approximately 0.207879576...[1]

### Factorial

The factorial of the imaginary unit i is most often given in terms of the gamma function evaluated at 1 + i:

$i! = \Gamma(1+i) \approx 0.4980 - 0.1549i.$

Also,

$|i!| = \sqrt{\pi \over \sinh(\pi)} \approx 0.521564... .$[2]

### Other operations

Many mathematical operations that can be carried out with real numbers can also be carried out with i, such as exponentiation, roots, logarithms, and trigonometric functions.

A number raised to the ni power is:

$\!\ x^{ni} = \cos(\ln(x^n)) + i \sin(\ln(x^n)).$

The nith root of a number is:

$\!\ \sqrt[ni]{x} = \cos(\ln(\sqrt[n]{x})) - i \sin(\ln(\sqrt[n]{x})).$

The imaginary-base logarithm of a number is:

$\log_i(x) = {{2 \ln(x)} \over i\pi}.$

As with any complex logarithm, the log base i is not uniquely defined.

The cosine of i is a real number:

$\cos(i) = \cosh(1) = {{e + 1/e} \over 2} = {{e^2 + 1} \over 2e} \approx 1.54308064... .$

And the sine of i is purely imaginary:

$\sin(i) = \sinh(1) \, i = {{e - 1/e} \over 2} \, i = {{e^2 - 1} \over 2e} \, i \approx 1.17520119 \, i... .$
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## Alternative notations

• In electrical engineering and related fields, the imaginary unit is often denoted by j to avoid confusion with electrical current as a function of time, traditionally denoted by i(t) or just i. The Python programming language also uses j to denote the imaginary unit. MATLAB associates both i and j with the imaginary unit, although 1i or 1j is preferable, for speed and improved robustness.[3]
• Some sources define[citation needed]j = −i, in particular with regard to traveling waves (e.g., a right traveling plane wave in the x direction) $e^{ i (kx - \omega t)} = e^{ j (\omega t-kx)} \,$.
• Some texts use the Greek letter iota (ι) for the imaginary unit, to avoid confusion, esp. with index and subscripts.
• Each of i, j, and k is an imaginary unit in the quaternions. In bivectors and biquaternions an additional imaginary unit h is used.
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## Matrices

When 2 × 2 real matrices m are used for a source, and the number one (1) is identified with the identity matrix, and minus one (−1) with the negative of the identity matrix, then there are many solutions to m2 = −1. In fact, there are many solutions to m2 = +1 and m2 = 0 also. Any such m can be taken as a basis vector, along with 1, to form a planar algebra.

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## Notes

1. ^ To find such a number, one can solve the equations
(x + iy)2 = i
x2 + 2ixyy2 = i
Because the real and imaginary parts are always separate, we regroup the terms:
x2y2 + 2ixy = 0 + i
and get a system of two equations:
x2y2 = 0
2xy = 1
Substituting into the first equation, we get
x2 − 1/4x2 = 0
x2 = 1/4x2
4x4 = 1
Because x is a real number, this equation has two real solutions for x: x = 1/2 and x = −1/2. Substituting both of these results into the equation 2xy = 1 in turn, we will get the same results for y. Thus, the square roots of i are the numbers 1/2 + i/2 and −1/2i/2. (University of Toronto Mathematics Network: What is the square root of i? URL retrieved March 26, 2007.)
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## References

1. ^ "The Penguin Dictionary of Curious and Interesting Numbers" by David Wells, Page 26.
2. ^ "abs(i!)", WolframAlpha.
3. ^
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