Math, imaginary
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i⁻³ = i¹ = i⁵ = i⁹ = +i
i⁻² = i² = i⁶ = i¹⁰ = –1
i⁻¹ = i³ = i⁷ = i¹¹ = –i
i⁰ = i⁴ = i⁸ = i¹² = +1
i^(4n) = 1 (n is integer)
i^(4n + 1) = i
i^(4n + 2) = -1
i^(4n + 3) = -i
√i = ±(√2/2)(1+i)
√(x+iy) = √[ (r+x)/2 ] ± i √[ (r–x)/2 ]
with r = √(x² + y²)

∛i = –i, ½+√3/2, –½+√3/2
∛(–1) = –1, ½–i√3/2, and ½–i√3/2
∛(–N) = ∛N[ –1, ½–i√3/2, and ½–i√3/2 ]

de Moivre's formula, states that for any complex number
(and, in particular, for any real number) x and
integer n it holds that"
(cos x + i sin x)ⁿ = cos (nx) + i sin (nx)
i = √(–1)

The property √(ab) = √a √b is valid when a and b
              are both non-negative real numbers.

Euler's formula is a mathematical formula in complex
analysis that establishes the deep relationship between
the trigonometric functions and the complex exponential
function. Euler's formula states that,
for any real number x:
e^(ix) = cos x + i sin x

Euler's Identity
e^(iπ) + 1 = 0
i = √(–1)
e is Euler's number, the base of natural logarithms
π is the ratio of the circumference of a circle to its
  diameter.

Derive square root of i
One way to obtain this answer is to solve the equation
(a+bi)=√i for a and b, or (a+bi)²=i
If you expand this equation using the
rules for complex multiplication, you get
(a²–b²) + (2ab)i = 0+i.
Equating real and imaginary parts gives you
a²–b²=0 and 2ab=1.

The equation a²–b²=0 means a=±b. However, if you plug
a=–b into the second equation you get –2b²=1 which can
not be satisfied by any real number b. Therefore, the
case a=–b is not possible, meaning a must equal b. Then
the second equation becomes 2a²=1. This means either
a=b=1/√2 or a=b=–1/√2.
That gets you the two square roots of i:
(1/√2)(1+i) and (–1/√2)(1+i)

Cube root
∛1 = 1, –(1/2) + i√3/2,  –(1/2) – i√3/2