User:Grover cleveland/complex

Real integer exponents
Exponentiation for real numbers satisfies the equations
 * $$x^{1} = x$$
 * $$x^{2} = x \cdot x$$
 * $$x^{3} = x \cdot x \cdot x$$

and generally:
 * $$x^{y + z} = x^{y} \cdot x^{z}$$

Applying this to the definitions above of multiplication for complex number in polar form, we may derive de Moivre's formula for any nonzero complex number z and any integer n:


 * $$z^{n} = |z|^{n} \cdot (\operatorname{cis} (n \cdot \arg(z)))$$

Real rational exponents
We may attempt to extend de Moivre's formula to a real fractional exponents. For example:


 * $$-1^{\frac{1}{2}} = (|-1|)^{\frac{1}{2}} \cdot (\operatorname{cis} (\frac{1}{2} \cdot \arg(-1)))$$
 * $$ = 1^{\frac{1}{2}} \cdot (\operatorname{cis} (\frac{1}{2} \cdot \pi))$$
 * $$ = 1 \cdot (i)$$
 * $$ = i$$

This seems satisfactory, in that $$i$$ satisfies the equation
 * $$i^{2} = -1$$.

However, recall that we initially defined $$\arg(z)$$, somewhat arbitrarily, to fall within the range
 * $$-\pi < \arg(z) \leq \pi$$

Had we instead chosen the range


 * $$-\pi \leq \arg(z) < \pi$$

then $$\arg(-1)$$ would be $$-\pi$$ rather than $$\pi$$, and$$-1^{\frac{1}{2}}$$ would come out as $$-i$$ rather than $$i$$.

Since $$-i$$ also satisfies the equation
 * $$(-i)^{2} = -1$$

it appears that our initial, somewhat arbitrary, choice about the range of $$\arg({z})$$ has had equally arbitrary consequences about the value of $$-1^{\frac{1}{2}}$$

Similarly, $$-1^{\frac{1}{3}}$$

Real integer exponents
By the normal definition of exponentiation:
 * $$x^{1} = x$$
 * $$x^{n+1} = x^{n} . x$$

Applying this to the formula for multiplication in polar form given above, we have de Moivre's formula, for any integer $$n$$:
 * $$ (r \operatorname{cis} \varphi)^{n} = r^{n} \operatorname{cis} (\varphi n) $$

It will be noted that, where $$n$$ is an integer, the choice of \varphi makes no difference to the result.

Real rational exponents
For a rational exponent expressed in lowest terms as $$\frac{p}{q}$$, then there are$$q$$ possible values for the result of exponentiation:
 * $$ (r \operatorname{cis} \varphi )^{ \frac{p}{q} } = $$