Chapter 2: Special Functions Flashcards
(25 cards)
exp(z)
as a power series defined everywhere on C
As a limit
And more importantly….
For z= x+iy
Exp(z) = sum from 0 to ∞ ( z^n / (n!))
Or
= limit as n tends to ∞ ( 1 + (z/n))^n
And so ( 1+ (x/n))^n tends to e^x as n tends to ∞ OR
Exp (z) = exp(x) (cosy * i siny)
Theorem 2.2
Properties of the exponential function
- if z=x+iy then |e^z| = e^x = e^{Re(z)}
- ∀z,w ∈C, e^{z+w} = e^z • e^w. (Sum powers if product)
- the exponential functi n e^z is PERIODIC with period 2pii
• ∀z ∈C. Exp(z) not equal to 0 and
1/ e^z = e^-z
Remember the real part of z is not equal to the modulus of z and in the exponential we have:
for all z in complex plane , e^z ≠0, 1/e^z = e^-z
Cosh z
Even function
Cosh z = 0.5( e^z + e^-z)
Defined on the complex plane
Cos z
Even function cos z = cos-z
Cos z = 0.5( e^iz + e^-iz)
Imaginary exponent power
Defined on the complex plane
Sinh z
Odd function
Sinh z = 0.5( e^z - e^-z)
Defined on the complex plane
Sin z
Odd function
Sin z = -sin z
Sin z = (1/2i) •( e^iz -e^-i z)
Defined on the complex plane
Relationships between cos sin sinh and cosh
- cos z = cosh(iz)
- sin (z+w) = sinz•cosw + cosz• sinw
- isinZ = sinh(iz)
- sin ² z + cos ² z = 1
This does not imply |sinz| ≤ 1. & |cosz| ≤ 1 for all complex z
(In fact sin and cos are unbounded on C)
Eg z=iy
|cosz| =
For y in reals
|cosz| = |cosiy| = |coshy| = coshy which tends to infinity as y tends to infinity
Example find M st
For all z with |z| =1
(e^z +cosz) /(6+z)| ≤ M
Let z=x+iy then
|cosz| = | 0.5(e^iz + e^-iz)|
By the triangle inequality
≤ 0.5|e^iz| + 0.5|e^-iz|
= 0.5 e^(Re(iz)) + 0.5 e^ (Re(-iz))
= 0.5 e^-y + 0.5 e^y = coshy
By definition of coshy
Hence for all z with |z|=1 |cosz| ≤ coshy ≤cosh1 and the triangle rule |e^z + cosz| ≤ |e^z| + coshy ≤ e^Re(z) + coshy ≤ e^1 +cosh1
|6+z| ≥ |6| -|z| =5 giving M = (e +cosh1)/(5)
Find the 0s of cosz
Let z = x+iy
(Remember z is in the complex plane so we have more roots)
Cosz = cos( x+iy)
= cos x cosiy - sinx siniy
By definition
= cosx (coshy) -i sinx(sinhy)
Taking the modulus and squaring:
|cosz| ² = cos ²x cosh ²y + sin ²x sinh ²y
(Note it’s a+ not - because the modulus doesn’t involve squaring I’s only the y part)
= cos ²x + sinh ²y
So we have
cos ²x + sinh ²y =0 ie (cosx-sinhy)(cosx +sinhy)=0
Cosx= -sinhy ——>
cos x =sinhy =0
Since x and y in reals
x= pi/2 + n*pi And y=0.
Thus solutions are
Z= pi/2 + n*pi (n=0, ±1, ±2,…) which are on the real axis
Tan z.
Excluding points at which the denominator is 0
Tan z = sinz /cosz
Except for z is (2n+1) π /2
arctan for arguement could be plus pi!!
Sec z
Excluding points at which the denominator is 0
Secz = 1 /cosz
Except for z is (2n+1) π /2
Tanhz
Excluding points at which the denominator is 0
Tanh z = sinhz /coshz
Except for z is (2n+1) πi /2
Cot z
Excluding points at which the denominator is 0
Cot z = cosz /sinz
Except for z is nπ
For integers n
Cosec z
Excluding points at which the denominator is 0
Cosec z= 1/sinz
Except for z is nπ
For integers n
Coth z
Excluding points at which the denominator is 0
Cothz = coshz /sinhz
Except for z is nπi
Cosech z
Excluding points at which the denominator is 0
Cosech z = 1/ sinhz
Except for z is nπi
properties Log Z
If log z has a value Then it has an infinite number of values and log z is not a function
•e^x is a strictly increasing positive function on R, define inverse function lnx on (0, ∞)
- this for all real x e^lnx =x
- if we suppose z is not equal to 0 any w in C st e^w = z is value of log z then for all n in Z
exp( w + 2nπi) = exp{w} = z
So if w is a value of log z so is w+ 2nπi is also, due to periodicity
log z is defined on C{0}
Example:
FInd all values of log(-e)
mod arg form: -e = e(cos π + isin π)
so a value of log(-e) is ln | − e| + i arg(−e) = 1 + iπ
All values are therefore given by:
1 + i(π + 2nπ) (n ∈ Z)
Example:
FInd all values of log( √3 − i)
mod arg form: √3 − i = 2(cos (-π/6) + isin(- π/6))
so a value of log( √3 − i) is
ln |2| + i arg(√3 − i) = ln 2 − iπ/6
All values are therefore given by:
ln2 - i(π/6) + (2nπi) (n ∈ Z)
VALUES OF LOG Z
complex logarithm
,if z ≠ 0 and z = reiθ with r bigger than 0, then
log z = ln r + iθ + 2nπi
= ln |z| + i arg z,
where arg z has infinitely many values, so that log z also has infinitely many values and
any pair differ by an integral multiple of 2πi
log z is not defined when z = 0
Find all roots of the equation cosh z = −i.
cosh z =(1/2)(e^z + e^−z)
so equation becomes
eᶻ + e⁻ᶻ = 2i
e^(2z) + 2ie^z + 1 = 0
quadratic eq with roots
eᶻ = -i ±√ ̅ ̅i2̅ ̅−̅ ̅1̅ = -i ± √2
by the quadratic formula
Thus required values of z are the values of log(−i ± i√2).
log(−i − i√2) has values ln(√2 + 1) −πi/2 + 2nπi (n ∈ Z),
log(−i + i√2) has values ln(√2 - 1) +πi/2 + 2nπi (n ∈ Z),
(ln(√2 − 1) = − ln(√2 + 1) )
REQUIRED SOLS ARE
z = ±[ln(√2 + 1) −πi/2 + 2nπi]
(n ∈ Z).
cosh^2y-sinh^2y
=1
EXAMPLE: Find all roots of e^2z +1+i =0
e^2z = -1-i
so values log(-1-i) are
ln√2 - 3πi/4 +2nπi n in Z
so z= 0.5 ln√2 - 3πi/8 + nπi n in Z
