Complex Numbers 5 6 7 Handout

8/13/2019 Complex Numbers 5 6 7 Handout

1/12

Complex exponentialTrigonometric and hyperbolic functions

Complex logarithmComplex power function

Chapter 13: Complex NumbersSections 13.5, 13.6 & 13.7

Chapter 13: Complex Numbers


2/12



DenitionProperties

1. Complex exponential

The exponential of a complex number z = x + iy is dened as

exp(z ) = exp( x + iy ) = exp( x ) exp( iy )= exp( x ) (cos( y ) + i sin(y )) .

As for real numbers, the exponential function is equal to itsderivative, i.e.

d

dz exp(z ) = exp( z ). (1)

The exponential is therefore entire.

You may also use the notation exp(z

) = e z

.



3/12



DenitionProperties

Properties of the exponential functionThe exponential function is periodic with period 2 i : indeed,for any integer k Z ,

exp(z + 2 k i ) = exp( x ) (cos( y + 2 k ) + i sin(y + 2 k ))= exp( x ) (cos( y ) + i sin(y )) = exp( z ).

Moreover,

| exp(z )| = | exp(x )| | exp(iy )| = exp( x ) cos2(y ) + sin 2(y )= exp( x ) = exp ( e (z )) .

As with real numbers,exp(z 1 + z 2) = exp( z 1) exp(z 2);

exp(z ) = 0.



4/12



Trigonometric functionsHyperbolic functions

2. Trigonometric functions

The complex sine and cosine functions are dened in a waysimilar to their real counterparts,

cos(z ) = e iz + e iz

2 , sin(z ) =

e iz e iz

2i . (2)

The tangent, cotangent, secant and cosecant are dened asusual. For instance,

tan( z ) = sin(z )cos(z )

, sec(z ) = 1cos(z )

, etc.



5/12




Trigonometric functions (continued)

The rules of differentiation that you are familiar with stillwork.

Example:Use the denitions of cos(z ) and sin(z ),

cos(z ) = e iz + e iz

2 , sin(z ) =

e iz e iz

2i .

to nd (cos(z )) and (sin(z )) .

Show that Eulers formula also works if is complex.



6/12




3. Hyperbolic functionsThe complex hyperbolic sine and cosine are dened in a waysimilar to their real counterparts,

cosh(z ) = e z + e z

2 , sinh(z ) =

e z e z

2 . (3)

The hyperbolic sine and cosine, as well as the sine and cosine,are entire.

We have the following relations

cosh(iz ) = cos( z ), sinh(iz ) = i sin(z ),(4)

cos(iz ) = cosh( z ), sin(iz ) = i sinh(z ).



7/12



DenitionPrincipal value of ln( | z | )

4. Complex logarithmThe logarithm w of z = 0 is dened as

e w = z .

Since the exponential is 2 i -periodic, the complex logarithm ismulti-valued.

Solving the above equation for w = w r + iw i and z = re i

gives

e w = e w r e iw i = re i = e w r = r w i = + 2p ,

which implies w r = ln( r ) and w i = + 2p , p Z .

Therefore,ln(z ) = ln( |z |) + i arg(z ).



8/12




Principal value of ln(z

)

We dene the principal value of ln(z ), Ln(z ), as the value of

ln(z ) obtained with the principal value of arg(z ), i.e.

Ln(z ) = ln( |z |) + i Arg(z ).

Note that Ln( z ) jumps by 2 i whenone crosses thenegative real axisfrom above.

y

x 0

Arg( z )=

1

1

Arg( z )> -

The negative real axis is called a branch cut of Ln(z ).



9/12




Principal value of ln(z

) (continued)

Recall thatLn(z ) = ln( |z |) + i Arg(z ).

Since Arg(z ) = arg( z ) + 2 p , p Z , we therefore see thatln(z ) is related to Ln(z ) by

ln(z ) = Ln( z ) + i 2p , p Z .

Examples:

Ln(2) = ln(2), but ln(2) = ln(2) + i 2p , p Z .Find Ln( 4) and ln( 4).

Find ln(10 i ).



10/12




Properties of the logarithmYou have to be careful when you use identities like

ln(z 1z 2) = ln( z 1)+ln( z 2), or ln z 1z 2= ln( z 1) ln(z 2).

They are only true up to multiples of 2 i .

For instance, if z 1 = i = exp( i / 2) and z 2 = 1 = exp( i ),

ln(z 1) = i 2

+ 2 p 1i , ln(z 2) = i + 2 p 2i , p 1, p 2 Z ,

andln(z 1 z 2) = i

32

+ 2 p 3i , p 3 Z ,

but p 3 is not necessarily equal to p 1 + p 2.Chapter 13: Complex Numbers


11/12




Properties of the logarithm (continued)Moreover, with z 1 = i = exp( i / 2) and z 2 = 1 = exp( i ),

Ln(z 1) = i 2 , Ln(z 2) = i ,

andLn(z 1 z 2) = i

2

= Ln( z 1) + Ln( z 2).

However, every branch of the logarithm (i.e. each expressionof ln(z ) with a given value of p Z ) is analytic except at thebranch point z = 0 and on the branch cut of ln( z ). In thedomain of analyticity of ln(z ),

d dz

(ln(z )) = 1z . (5)


C l i l


12/12



Denition

5. Complex power function

If z = 0 and c are complex numbers, we dene

z c = exp ( c ln(z ))= exp ( c Ln(z ) + 2 pc i ) , p Z .

For c C , this is again a multi-valued function, and we denethe principal value of z c as

z c = exp ( c Ln(z ))


Documents

Complex Numbers 5 6 7 Handout