EULER’S THEOREM
PREPARED BY: ENGR. MARVIN Y. VILLORENTE
Definition of eulers
cosx + jsin x =
𝒆𝒋𝒙 +𝒆−𝒋𝒙
𝟐
+
𝒆𝒋𝒙 −𝒆−𝒋𝒙
𝒋
𝒋𝟐
TRIGONOMETRIC FUNCTION OF COMPLEX
NUMBERS
1. Sin x
𝑒 𝑗𝑥 −𝑒 −𝑗𝑥
=
𝑗2
2. Cos x=
3.
4.
𝑒 𝑗𝑥 +𝑒 −𝑗𝑥
2
𝑒 𝑗𝑥 −𝑒 −𝑗𝑥
Tanx =−j 𝑗𝑥 −𝑗𝑥
𝑒 +𝑒
𝑒 𝑗𝑥 +𝑒 −𝑗𝑥
Cotx= j 𝑗𝑥 −𝑗𝑥
𝑒 −𝑒
𝑗𝑥
5. Secx=1/cosx=2/𝑒 + 𝑒 −𝑗𝑥
6. Cscx=1/sinx=j2/𝑒 𝑗𝑥 − 𝑒 −𝑗𝑥
tanx=sinx/cosx =
𝑒𝑗𝑥 −𝑒−𝑗𝑥
𝑗2
𝑒𝑗𝑥 +𝑒−𝑗𝑥
2
=
𝑒 𝑗𝑥 −𝑒 −𝑗𝑥 𝑗
𝑗(𝑒 𝑗𝑥 +𝑒 −𝑗𝑥 ) 𝑗
INVERSE TRIGO FUNCTION OF COMPLEX
NUMBERS
1. Arcsin x = −jln(jx ± 1 − 𝑥 2 )
2. Arccosx=−jln(x ± 𝑥 2 − 1)
3. Arctanx=−jln
1+𝑗𝑥
1−𝑗𝑥
4. Arccotx=−jln
𝑥+𝑗
𝑥−𝑗
5.
6.
1± 1−𝑥 2
Arcsecx=−jln
𝑥
𝑗± 𝑥 2 −1
Arccscx=−jln
𝑥
-j ln(𝒋𝒙 ± 𝟏 − 𝒙𝟐 )
𝑗
2
𝑗−𝑥 𝑗
ln
.
𝑗+𝑥 𝑗
𝑗 2 −𝑗𝑥 −1−𝑗𝑥
=2 =
𝑗 +𝑗𝑥 −1+𝑗𝑥
1
−
2
=
−1
2
=
=
1
−2
lnab=blna
−(1+𝑗𝑥) (1+𝑗𝑥)
=
−1+𝑗𝑥
1−𝑗𝑥
Derivation:
Let y=Arcsin x
Siny=x
𝒆𝒋𝒚 −𝒆−𝒋𝒚
x=
𝒋𝟐
𝒆𝒋𝒚 −𝟏 𝒋𝒚
x=
𝒆
𝒋𝟐
x2+2x=y
x2+2x +1 =y +1
(x+1)2=y+1
(x2)3=x6
(22)3=26
43=26
64=64
x𝒆𝒋𝒚 (𝒋𝟐)=𝒆𝟐𝒋𝒚 − 𝟏
𝒆𝟐𝒋𝒚 - 2x𝒆𝒋𝒚 j=1
𝒆𝟐𝒋𝒚 −𝟏
x=
𝒆𝒋𝒚
𝒋𝟐
𝒆𝟐𝒋𝒚 −𝟏
x= 𝒋𝒚
𝒆 (𝒋𝟐)
𝒆𝟐𝒋𝒚 - 2x𝒆𝒋𝒚 j+ (𝒋𝒙)𝟐 = 𝟏 + (𝒋𝒙)𝟐
(𝒆𝒋𝒚 − 𝒋𝒙)𝟐 = 𝟏 − 𝒙𝟐
(𝒆𝒋𝒚 − 𝒋𝒙)𝟐 = 𝟏 − 𝒙𝟐
𝒆𝒋𝒚 − 𝒋𝒙=± 𝟏 − 𝒙𝟐
𝒆𝒋𝒚 =𝒋𝒙 ± 𝟏 − 𝒙𝟐 ln both side
jy=ln(𝒋𝒙 ± 𝟏 − 𝒙𝟐 )
ln(𝒋𝒙± 𝟏−𝒙𝟐 ) 𝑗
y=
𝑗
𝑗
y=arcsinx= -j ln(𝒋𝒙 ± 𝟏 − 𝒙𝟐 ) formula
Let y=Arctanx
tan y= x
𝑒 𝑗𝑦 −𝑒 −𝑗𝑦
x=−j 𝑗𝑦 −𝑗𝑦
𝑒 +𝑒
𝑒 𝑗𝑦 −1 𝑗𝑦
x=−j
𝑒
𝑒 𝑗𝑦 +1 𝑗𝑦
𝑒
𝑒2𝑗𝑦 −1
x=-j𝑒2𝑗𝑦+1
𝑒𝑗𝑦
𝑒𝑗𝑦
𝑒 2𝑗𝑦 −1
x=−j 2𝑗𝑦
𝑒
+1
2𝑗𝑦
x(𝑒
+1)=-j(𝑒 2𝑗𝑦 − 1)
𝑥𝑒 2𝑗𝑦 + 𝑥 = −𝑗𝑒 2𝑗𝑦 + 𝑗
𝑥𝑒 2𝑗𝑦 +𝑗𝑒 2𝑗𝑦 = 𝑗 − 𝑥
𝑒 2𝑗𝑦 𝑥 + 𝑗 = 𝑗 − 𝑥
𝑒 2𝑗𝑦
=
𝑗−𝑥
𝑥+𝑗
ln both side
𝑗−𝑥
2jy= ln
𝑗+𝑥
1
𝑗−𝑥
y = ln
𝑗2 𝑗+𝑥
−𝑗 𝑗−𝑥
y= ln
-formula
2
𝑗+𝑥
example
Evaluate the ff and express the result in polar form
1. arcsin(3+j4)
2. Arccos(4-j5)
3. Actan(2-j3)
1. arcsin(3+j4) = −jln(j(3+j4) ± 1 − (3+j4)2 )
= −jln(j(3+j4) ± 8 − 𝑗24)
=−jln(j(3+j4) ± (25.3∠ − 71.6)1/2 )
= −jln(j(3+j4) ± 5.03∠ − 35.8)
Use +:
=−jln(j 3+j4 + 5.03∠ − 35.8)
= - jln(0.079651+j0.0576624)
=-jln(0.09833ej0.6266)
=-jln(0.09833ej0.6266)
= -j(ln(0.09833)+lnej0.6266)
=-j(-2.31943+j0.6266)
z1=0.6266+j2.31943
z1=2.4026∠74.88 ans
Use (-):
= −jln(j 3+j4 − 5.03∠ − 35.8)
=−jln(−8.0797 + j5.94234)
=-jln(10.03ej2.5075)
=-j(ln10.03+lnej2.5075)
=-j(ln10.03+lnej2.5075)
= -j(2.3056+j2.5075)
= 2.5075-j2.3056
z2=3.4064∠-42.6 ans.