De Moivre's formula

La De Moivre formula, so named by Abraham de Moivre states that for any complex number (and in particular for any real number) ${displaystyle x}$ and for any ${displaystyle nin mathbb {Z} }$ verified that

${displaystyle (cos(x)+ioperatorname {sen}(x))^{n}=cos(nx)+ioperatorname {sen}(nx)}}$.

This formula connects complex numbers (i stands for imaginary unit) with trigonometry.

The expression ${displaystyle cos x+ioperatorname {sen} x}$ sometimes abbreviates as ${displaystyle operatorname {cis} x}$.

By expanding the left side of equality and comparing the real part with the imaginary, it is possible to obtain very useful expressions for ${displaystyle cos(nx)}$ and ${displaystyle operatorname {sen}(nx)}$ in terms of ${displaystyle cos x}$ and ${displaystyle operatorname {sen} x}$. In addition, this formula can be used to find explicit expressions for the ${displaystyle n}$- that's the root of the unit, that's, complex numbers. ${displaystyle z}$ such as ${displaystyle z^{n}=1}$.

History

Seal with the effigy of Euler

The current form of the formula appears in the work Introductio in analysin infiniterumof Euler, which demonstrates it for all natural integers ${displaystyle n}$ in 1748. But it also appears implicitly in the works of Abraham de Moivre several times since 1707, in his work on the roots ${displaystyle n}$-thousands of complex numbers. In fact, the two problems are related: writing that (gasps) x + i without x)ⁿ = cos(nx+ i sin()nx) is equivalent to saying that # x + i without x it is one of the unique roots of the complex cos(nx+ i sin()nx).

Relation to Euler's formula

Moivre's formula can be obtained from Euler's formula:

${displaystyle e^{ix}=cos x+ioperatorname {sen} x}$

applying laws of exponentiation

${displaystyle left(e^{ix}right)^{n}=e^{inx}$

Then, by Euler's formula,

${displaystyle e^{i(nx)}=cos(nx)+ioperatorname {sen}(nx}$.

Some results

Starting again from Euler's formula:

${displaystyle e^{ix}=cos x+ioperatorname {sen} x}$

If we do ${displaystyle x=pi }$ Then we have the identity of Euler:

${displaystyle {begin{aligned}e^{ipi } alien=cos pi +isin pi \ fake=-1+0{aligned}}}}}}}$

That is:

${displaystyle e^{ipi }=-1,}$

Furthermore, since we have these two equalities:

${displaystyle e^{ix}=cos x+ioperatorname {sen} x}$

${displaystyle e^{-ix}=cos x-ioperatorname {sen} x}$

we can deduce the following:

${displaystyle {begin{aligned}cos x fake={frac {e^{ix}{-ix}{2}}{2}{operatortorname {sen} x fake={frac {e^{ix}-e^{ix}}}{2i}}}}{aligned}}}}}}}}$

Proof by induction

We consider three cases.

For an integer ${displaystyle n/2003/0}$We came from mathematical induction. When ${displaystyle n=1}$ the result is clearly true. For our hypothesis we assume that the result is true for some positive integer ${displaystyle k}$. That's what we assume:

${displaystyle left(cos x+ioperatorname {sen} xright)^{k}=cos(kx)+ioperatorname {sen}(kx)}}$

Now, considering the case ${displaystyle n=k+1}$:

$[sighs]$

We deduce that the result is true for n = k + 1 when it is true for n = k. By the principle of mathematical induction it follows that the result is true for all positive integers n≥1.

When ${displaystyle n=0}$ the formula is true since ${displaystyle cos(0x)+ioperatorname {sen}(0x)=1+i0=1}$and (by convention) ${displaystyle z^{0}=1}$.

When ${displaystyle n vis0}$, we believe that there is a positive integer ${displaystyle m}$ such as ${displaystyle n=-m}$, so

${cHFFFFFF}{cHFFFFFF}{cHFFFFFFFF}{cHFFFFFFFF}{cHFFFFFFFF}{cHFFFFFFFFFFFFFFFFFFFFFFFFFF}{cHFFFFFFFFFFFF}{cHFFFFFFFF}{cHFFFFFFFFFFFF}{x}{cHFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF}{c}{c}{cHFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF}{c}{cHFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF}{c}{$

Therefore theorem is true for all ${displaystyle nin mathbb {Z} }$.

Generalization

The formula is actually true in a field much more general than the one presented above: yes z and w are complex numbers, then

${displaystyle left(cos z+ioperatorname {sen} zright)^{w}}}$

is a multivalued function while

${displaystyle cos(wz)+ioperatorname {sen}(wz)}$

don't be. Therefore it can be ensured that:

${displaystyle cos(wz)+ioperatorname {sen}(wz)}$is a value of${displaystyle left(cos z+isin zright)^{w},}$.

Applications

This formula can be used to find both the power and the nth roots of a complex number written in polar form.

${displaystyle z=rleft(cos x+ioperatorname {sen} xright)}$

If the complex number is in binary form, first it must be converted to polar form, being ${displaystyle r}$ the module.

Power

To obtain the power of the complex number, the formula is applied:

${displaystyle z^{n}=left[left(cos(x)+ioperatortorname {sen}(x)right)right]^{n}= viszficient^{n}left[cos(nx)+ioperatorname {sen}(nx)right]}}}}$

Roots

To get the ${displaystyle n}$ roots of a complex number, apply:

${displaystyle z^{1/n}=left[rleft(cos x+ioperatorname {sen} xright)right]^{1/n}=r^{1/n}left[cos left({cos left({x+2kpi }{n}{nright}{s}{s}{s}{s}{s}{s}{s}{s}{s}{s}{s}{cccccs}{s}{s}{cs}{s}{ccnright(s)}{s}{s}{s}{s)}{s}{s}{s}{s}{s}{s}{s}{cs)}$

where ${displaystyle k}$ It's an integer that goes from ${displaystyle}$ until ${displaystyle n-1}$that by replacing it in the formula allows to obtain ${displaystyle n}$ different roots ${displaystyle z}$.