Imaginary unit

( x, y ) is confined by hyperbola xy = –1 for an imaginary unit matrix.

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

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

and

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

would be solutions to the matrix equation

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

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, R) has exactly two elements: The identity and the automorphism which exchanges "CW" (clockwise) and "CCW" (counter-clockwise) rotations. For more, see orthogonal group.

All these ambiguities can be solved by adopting a more rigorous definition of complex number, and by 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.

Consider the matrix equation

$${\displaystyle {\begin{pmatrix}z&x\\y&-z\end{pmatrix}}^{2}={\begin{pmatrix}-1&0\\0&-1\end{pmatrix}}.}$$

Here, z² + xy = –1, so the product xy is negative because xy = –(1 + z²), thus the point (x, y) lies in quadrant II or IV. Furthermore,

$${\displaystyle z^{2}=-(1+xy)\geq 0\implies xy\leq -1}$$

so (x, y) is bounded by the hyperbola xy = –1.

The two square roots of i in the complex plane

The three cube roots of i in the complex plane

Just like all nonzero complex numbers, i has two square roots: they are[lower-alpha 2]

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

Indeed, squaring both expressions yields:

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

Using the radical sign for the principal square root, we get:

${\displaystyle {\sqrt {i}}={\frac {\sqrt {2}}{2}}(1+i).}$

The three cube roots of i are:

${\displaystyle -i,}$

${\displaystyle {\frac {\sqrt {3}}{2}}+{\frac {i}{2}},}$ and

${\displaystyle -{\frac {\sqrt {3}}{2}}+{\frac {i}{2}}.}$

Similar to all the roots of 1, all the roots of i are the vertices of regular polygons, which are inscribed within the unit circle in the complex plane.

Multiplying a complex number by i gives:

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

(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:

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

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

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

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

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

${\displaystyle i^{4n}=1}$

Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle i^{4n+1} = i}

${\displaystyle i^{4n+2}=-1}$

${\displaystyle i^{4n+3}=-i,}$

This leads to the conclusion that

${\displaystyle i^{n}=i^{(n{\bmod {4}})}}$

where mod represents the modulo operation. Equivalently:

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

Making use of Euler's formula, iⁱ is

${\displaystyle i^{i}=\left(e^{i(\pi /2+2k\pi )}\right)^{i}=e^{i^{2}(\pi /2+2k\pi )}=e^{-(\pi /2+2k\pi )}}$

where k ∈ Z, the set of integers.

The principal value (for k = 0) is e^−π/2, or approximately 0.207879576.[10]

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

${\displaystyle i!=\Gamma (1+i)\approx 0.4980-0.1549i~.}$

Also,

${\displaystyle |i!|={\sqrt {\frac {\pi }{\sinh \pi }}}}$[11]

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. All of the following functions are complex multi-valued functions, and it should be clearly stated which branch of the Riemann surface the function is defined on in practice. Listed below are results for the most commonly chosen branch.

A number raised to the ni power is:

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

The ni^th root of a number is:

${\displaystyle {\sqrt[{ni}]{x}}=\cos \left({\frac {\ln x}{n}}\right)-i\sin \left({\frac {\ln x}{n}}\right)~.}$

The imaginary-base logarithm of a number is:

${\displaystyle \log _{i}x={\frac {2\ln x}{i\pi }}~.}$

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

The cosine of i is a real number:

${\displaystyle \cos i=\cosh 1={\frac {e+1/e}{2}}={\frac {e^{2}+1}{2e}}\approx 1.54308064\ldots }$

And the sine of i is purely imaginary:

${\displaystyle \sin i=i\sinh 1={\frac {e-1/e}{2}}i={\frac {e^{2}-1}{2e}}i\approx (1.17520119\ldots )i~.}$