Copyright © 2022
e-ISSN: 2686-2573 DOI: 10.21512/emacsjournal.v4i1.7999
15 JURNAL EMACS
(Engineering, MAthematics and Computer Science) Vol.4 No.1 January 2022: 15-17
Euler Formula Derivation
Wikaria Gazali
Mathematics & Statistics Department, School of Computer Science, Bina Nusantara University,
Jakarta, Indonesia 11480 [email protected]
*Correspondence: [email protected]
Abstract – This paper discusses the derivation of Euler’s formula. To obtain this model, the writer derives Euler’s formula from ex+iy by first finding the norm and argument of ex+iy. In this derivation we substitute the norm and argument of ex+iy on complex numbers in polar coordinates, until we get the derivation of Euler’s formula.
Keywords: Euler; formula; norm; argument.
I. INTRODUCTION
Euler’s formula is widely used in solving mathematics or calculus, especially in solving complex numbers. Euler’s formula is also used in The Exact Iterative Riemann Solver, The Approximate Riemann Solver of Roe, The HLLE Riemann Solver. To obtain this model, the authors use the derivation from e
x 1 1
lim x=
+
∞
x→ .
In this derivation we substitute the norm and argument of ex+iy on complex numbers in polar coordinates, until we get the derivation of Euler’s formula..
II. METHODS
The inventor of the Euler formula is Leonhard Euler (1707 - 1783)
e
iθ =cos θ
+i sin θ
, but the derivation of the formula is rarely stated clearly by mathematicians, therefore the author will convey in detail where the origin of the Euler formula is, namely by finding the length of the vector ex+iy with the Limit method and the argument of ex+iy using complex numbers in Polar coordinates.III. RESULTS AND DISCUSSION
In this case, to get the derivation of Euler’s formula, the writer describes the norm and argument of ex+iy.
First we find the norm of ex+iy . We already know that e
x 1 1
lim x=
+
∞ x→
analog
n e 1 1
lim n=
+
∞ n→
,
so that e
n 1 x
lim n x
=
+
∞
n→ ,
or
n 1 x e lim
n
x
+
∞
= →
n .
Given that the above formula applies to the complex number z, then by substituting x = z,
we get
n 1 z
ez lim n
+
∞
= → n
,
where z = x + iy (complex numbers in Cartesian coordinates),
so that
n iy 1 x
ex iy lim n
+ +
∞
= →
+
n ,
e lim 1 nx
n iy
x
+
+
∞
= →
+
n iy
n ...(1)
16 JURNAL EMACS (Engineering, MAthematics and Computer Science) Vol.4 No.1 January 2022: 15-17 Norm from ex+iy is
n 1 x e lim
2 n 2 iy
x
+
+
∞
= →
+
n y
n ,
2 lim 1 e
n
2 2 2 iy 2
x 2
1
+ + +
∞
= →
+
n y n x n
x n
,
2 lim 1 e
n 2
2 iy 2
x
12
+ +
∞ +
= →
+
n y x n
x
n ,
( )
e n
lim 2 iy
x 2 12
2 2
+ +
∞
+ = → n
y x n
x
en ,
e 2
lim iy x
2 2
+ +
∞
+ = → n
y x x
en ,
e 2
lim iy x
2 2
∞ + +
∞ + = →
y x x
en ,
( )
e 0
lim iy
x + =en→∞x+ ,
So
e
x +iy= e
x...(2) To find the argument of ex+iy, we first consider complex numbers in Polar coordinates
) sin i
(cosθ+ θ
=r
z , then , such that
x tgy arc
θ = ,
and zn =rn(cosθ+ isin θ)n.
Where (cosθ+ isin θ)n =(cosnθ+ isin nθ) according to De Moivre’s Theorem,
then zn =rn(cosnθ + isin nθ), so that, or x
tgy n )
arg(zn = arc .
Then we determine the argument of ex+iy, where from equation (1)
we have
n 1 x e lim
n iy
x
+
+
∞
= →
+
n iy
n ,
then
+
∞
= →
+
n 1 x tg lim n
) e
arg( x iy n
y n arc
,
+
∞
= →
+
n n n y n arc
n 1 x tg lim n
) e
arg( x iy ,
+
∞
= →
+
x n arc y nlim n tg )
e arg( x iy
x n
y x
n yn x arc y
n +
+
∞ +
= →
+ tg
lim n ) e
arg( x iy .
Because
1 1 1 lim 1 1
1 1 1
1 lim 1
tg1 lim
2 '
'
2
=
∞ +
= →
+
∞
= →
∞
→
t t
t t t t
t arc t
t ,
then
x n
yn n→∞ +
+ )= lim
e
arg( x iy , or
) 1 e
arg( x +iy = y. So
arg( e
x +iy)
= y ...(3)Complex numbers in Polar coordinates
)
sin i
(cos θ
+θ
=r
z , where r= z and θ =arg
( )
z , so that z= z[
cos{
arg( )
z}
+ isin{
arg( )
z} ]
...(4)Let
z = e
x +iy, then equation (4) will change to:( )
{ } { ( ) }
[
cos arge isin arge]
e
ex +iy = x +iy x +iy + x +iy ...(5)
Substitute (2) and (3) in (5), so that equation (5) becomes:
) sin i (cos
e
x +iy =ex y+ y ,) sin i (cos e
ex iy =ex y+ y ,
So that
e
iy =cos
y+i sin
y...(6)Substitute y=θ in equation (6), then equation (6) changes to :
e
iθ =cos θ
+i sin θ
17 Euler Formula Derivation… (Wikaria Gazali)
IV. CONCLUSION
In this decrease, a decrease has been carried out by searching first.
The Norm of ex+iy is
e
x +iy= e
xand the argument of ex+iy is
arg( e
x +iy) = y
, then by substituting for complex numbers in polar coordinates is( )
{ } { ( ) } [ cos arg e i sin arg e ]
e
e
x +iy=
x +iy x +iy+
x +iyDerivation of the Euler’s formula has been described above and obtained its derivation :
θ θ
θ
cos i sin
e
i= +
where
x tg y arc
θ =
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