Solutions to Residue Problems:

1) Find the residue for f(z) =  3 exp (z)/  z ⁴

 

We know from the residue formula:

 

 Res f(z) = c _{- 1}    

 

= 1/ (m – 1)!  lim _{z ® a}    d ^{m - 1}   / dz ^{m - 1}    {(z – a _k) ^m    f(z) }

The pole is of order m = 4 at z = 0, then:

 

Res f(z) = 1/ 3!  lim _{z ® 0}      d ³   / dz³    {(z -  0) ⁴     ×  3 exp (z) / z ⁴  }

 

Res f(z) = 1/ 3!  lim _{z ® 0}     d ³   / dz ³   { 3exp (z)}

Res f(z) = 1/6   lim _{z ® 0}     { 3exp (z)} = 3/6  = ½

2) Find Res f(z) for cos z/ z ⁵

 

From the residue formula:      Res f(z) = c _{- 1}     

 

= 1/  (m – 1)!  lim _{z ® a}      d ^{m - 1}   / dz ^{m - 1}    {(z – a _k) ^m    f(z) }

 

The pole is of order m = 5 at z = 0  so that:

Res f(z) = 1/ 4!  lim _{z ® 0}      d ⁴   / dz⁴    {(z -  0) ⁵     ×  cos (z) / z ⁵ }

Res f(z) = 1/ 4!  lim _{z ® 0}     d ⁴   / dz ⁴   { cos  (z)}

 

Taking the 4^th derivative of cos z:

 

1) d/ dz (cos z) = - sin z and 2)  d/dz (-sin z) = - cos z

 

3) d/dz (- cos z) = - (-sin z) = sin z and  (4) d/dz (sin z) = cos z

 

Therefore:

 

Res f(z) =  1/ 4!  lim _{z ® 0}   (cos z) =  1/24 (1) = 1/ 24

3) Find all the residues at those singular points inside the circle  ÷ z  ÷     =    2  

 

For:  f(z) =   z ²   / ( z ⁴  - 1)

Rewrite f(z):

z ²  / ( z ⁴  - 1)  =    z ²   / ( z ²  - 1) ( z ²  +  1)

Look at factors of the denominator:

 

( z ²  +  1)  =   (z + i)  (z – i)

 

( z ²  - 1)    =   (z + 1) (z – 1)

 

Then rewrite f(z) again:

z ² / ( z ⁴  - 1)  =    =    z ²   / ( z  - 1) ( z  +  1) (z + i)  (z – i)

 

The order of the poles is simple, i.e.  m = 1  and 4 singular points occur inside the circle  ÷ z  ÷   =   2  

z= 1, z = -1, z = i, z = -i

 

Look at z = 1:

Res (f(z)) =  (z – a)     f(z)  = 


lim _{z ® 1}   (z – 1)   [z ²  / ( z  - 1) ( z  +  1) (z + i)  (z – i)]

    

Res (f(z)) =  lim _{z ® 1}   [z ²   / ( z  +  1) (z + i)  (z – i) ] =


  z ² / 4 z ³  ÷ _{z = 1}   = 1 / 4(1) ³   =  ¼

 

 

Note: by L’Hospital’s rule: 

 

Res [f, z_o] =  lim _{z ® z o}   1 / ( z ⁴  - 1)  =

lim _{z ® z o}   1  / 4 z ³     =  ( 1/ 4 z_o ³  )

Proceeding, for z = -1:

 

Res (f(z)) =  lim _{z ®  -1}   [z ²   / ( z  -  1) (z + i)  (z – i)] = 


  z ² / 4 z ³ ÷  _{z =- 1}      = (-1)²   / 4(-1) ³ =  - ¼

 

Similarly, for z = -i:

 

Res (f(z)) =  lim _{z ®  -i}   [z ²  / ( z  -  1) (z + 1)  (z – i) ]

 

=  z ² / 4 z ³ ÷  _z =- -i   =  (-i) ² / 4(-i) ³ =  -1/ 4i =  1/ 4i =  -i/ -4 = i/4

Similarly, for z = i:

 

Res (f(z)) =  lim _{z ®  i}   [z ²  / ( z  -  1) (z + 1)  (z + i)

=    z ² / 4 z ³ ÷  _{z = i}  =  (i) ² / 4(i) ³ =  -1/ -4i =  1/ 4i =  -i/ 4