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Section 2.4 Limits and Continuity

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COMPLEX ANALYSIS: Maple Worksheets,  2001
(c) John H. Mathews          Russell W. Howell
mathews@fullerton.edu     howell@westmont.edu

Complimentary software to accompany the textbook:

COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc.,      40  Tall  Pine  Drive,      Sudbury,  MA  01776
Tele.  (800) 832-0034;      FAX:  (508)  443-8000,      E-mail:  mkt@jbpub.com,      http://www.jbpub.com/


CHAPTER 2   COMPLEX FUNCTIONS

Section 2.4  Limits and Continuity

     Let  u = u(x, y)  be a real-valued function of the two real variables  x  and  y .  Recall that u has the u[0] as (x, y ) approaches (x[0], y[0] ) provided that the value of u(x, y) can be made to get as close as we please to the value u(x[0], y[0]) by taking (x, y ) to be sufficiently close to (x[0], y[0] ).  


Example 2.14, Page 69. The function  u(x, y) = 2*x^3/(x^2+y^2)  has the limit  0  as (x , y ) approaches (0 , 0 ) .

> t:='t': u:='u': x:='x': y:='y':
u := proc(x,y)  2*x^3/(x^2+y^2)  end:
`u(x,y) ` = u(x,y); ` `;
lim1 := limit(u(x,y), x=0):
lim2 := limit(lim1, y=0):
`limit  u(x,y)  as  x->0 ` = lim1; `and`;
`limit  u(x,y)  as  x->0 and y->0 ` = lim2;

`u(x,y) ` = 2*x^3/(x^2+y^2)

` `

`limit u(x,y) as x->0 ` = 0

and

`limit u(x,y) as x->0 and y->0 ` = 0

> `u(x,y) ` = u(x,y); ` `;
lim1 := limit(u(x,y), y=0):
lim2 := limit(lim1, x=0):
`limit  u(x,y)  as  y->0 ` = lim1; `and`;
`limit  u(x,y)  as  y->0 and x->0 ` = lim2;

`u(x,y) ` = 2*x^3/(x^2+y^2)

` `

`limit u(x,y) as y->0 ` = 2*x

and

`limit u(x,y) as y->0 and x->0 ` = 0

> U := subs({x=r*cos(t),y=r*sin(t)},u(x,y)):
`u(r cos t,r sin t) ` = U; ` `;
lim1 := limit(U, r=0):
`limit  u(r cos t,r sin t)  as  r->0 ` = lim1;

`u(r cos t,r sin t) ` = 2*r^3*cos(t)^3/(r^2*cos(t)^2+r^2*sin(t)^2)

` `

`limit u(r cos t,r sin t) as r->0 ` = 0

So, along all lines through the origin, the limit is  0 .


Example 2.15, Page 70.   The function  u(x, y) = x*y/(x^2+y^2)  
does NOT have a limit as  (x , y )  approaches  (0 , 0 ) .

> t:='t': u:='u': x:='x': y:='y':
u := proc(x,y)  x*y/(x^2+y^2)  end:
`u(x,y) ` = u(x,y); ` `;
lim1 := limit(u(x,y), x=0):
lim2 := limit(lim1, y=0):
`limit  u(x,y)  as  x->0 ` = lim1; `and`;
`limit  u(x,y)  as  x->0 and y->0 ` = lim2;

`u(x,y) ` = x*y/(x^2+y^2)

` `

`limit u(x,y) as x->0 ` = 0

and

`limit u(x,y) as x->0 and y->0 ` = 0

> `u(x,y) ` = u(x,y); ` `;
lim1 := limit(u(x,y), y=0):
lim2 := limit(lim1, x=0):
`limit  u(x,y)  as  y->0 ` = lim1; `and`;
`limit  u(x,y)  as  y->0 and x->0 ` = lim2;

`u(x,y) ` = x*y/(x^2+y^2)

` `

`limit u(x,y) as y->0 ` = 0

and

`limit u(x,y) as y->0 and x->0 ` = 0

> U := subs({x=r*cos(t),y=r*sin(t)},u(x,y)):
`u(r cos t,r sin t) ` = U; ` `;
lim1 := limit(U, r=0):
`limit  u(r cos t,r sin t)  as  r->0 ` = simplify(lim1);

`u(r cos t,r sin t) ` = r^2*cos(t)*sin(t)/(r^2*cos(t)^2+r^2*sin(t)^2)

` `

`limit u(r cos t,r sin t) as r->0 ` = cos(t)*sin(t)

Since this value is dependent on the angle of approach to  0 , u(x, y) = x*y/(x^2+y^2)
 does NOT have a limit  as  (x , y )  approaches  (0 , 0 ) .

Theorem 2.1   Let  f(z) = u(x, y)+i*v(x, y)  be a complex function that is defined in some neighborhood of  z[0] ,  

except perhaps at  z[0] = x[0]+i*y[0] .  Then  

     Limit(f(z), z = z[0])  =  w[0] = u[0]+i*v[0]  

if and only if

     Limit(u(x, y), `(x,y)` = (x[0], y[0])) = u[0]   and   Limit(v(x, y), `(x,y)` = (x[0], y[0])) = v[0] .  

  Limits of complex functions are formally the same as in the case of real functions, and the sum, difference, product, and quotient of functions have limits given by the sum, difference, product, and quotient of the respective limits. These proofs are left as exercises.  


Example 2.17, Page 73.   Find   limit(f(z), z = 1+i)   for   f(z) = z^2-2*z+1 .

> f:='f': z:='z':
f := z -> z^2 - 2*z + 1:
`f(z) ` = f(z); ` `;
`limit  f(z)  as  z->1+i ` = limit(f(z), z=1+I);
`Also, the value of f(1+i) is:`;
`f(1+i) ` = f(1+I);

`f(z) ` = z^2-2*z+1

` `

`limit f(z) as z->1+i ` = -1

`Also, the value of f(1+i) is:`

`f(1+i) ` = -1


Example 2.18, Page 75.   Show that the polynomial function given by
    P(z) = a[0]+a[1]*z+a[2]*z^2 + ... + a[n]*z^n
is continuous at each point  z[0]  in the complex plane.
For illustration, we use  n = 5 .

> P:='P': z:='z': z0:='z0':
P := z -> sum('a[k]'*z^k, 'k'=0..5):
`P(z) ` = P(z);
lim := limit(P(z), z=z0):
`limit P(z)  as  z->z0 ` = lim;
`Also, the value of P(z0) is:`;
`P(z0) ` = P(z0); ` `;
`P(z0) = limit P(z)  as  z->z0 `;
evalb(limit(P(z), z=z0) = P(z0));

`P(z) ` = a[0]+a[1]*z+a[2]*z^2+a[3]*z^3+a[4]*z^4+a[5]*z^5

`limit P(z) as z->z0 ` = a[0]+a[1]*z0+a[2]*z0^2+a[3]*z0^3+a[4]*z0^4+a[5]*z0^5

`Also, the value of P(z0) is:`

`P(z0) ` = a[0]+a[1]*z0+a[2]*z0^2+a[3]*z0^3+a[4]*z0^4+a[5]*z0^5

` `

`P(z0) = limit P(z) as z->z0 `

true

Example 2.19, Page 76.  Find   limit(f(z), z = 1+i)   for   f(z) = (z^2-2*i)/(z^2-2*z+2)  .

> f:='f': z:='z':
f := z -> (z^2 - 2*I)/(z^2 - 2*z + 2):
`f(z) ` = f(z);
fun := f(1+I):
`f(1+I) ` = undefined;
`However,`;
lim := limit(f(z) ,z=1+I):
`limit  f(z)  as  z->1+i ` = lim;

`f(z) ` = (z^2-2*I)/(z^2-2*z+2)

Error, (in f) numeric exception: division by zero

`f(1+I) ` = undefined

`However,`

`limit f(z) as z->1+i ` = 1-I

> f:='f': F:='F': z:='z': Z :='Z':
f := z -> (z^2 - 2*I)/(z^2 - 2*z + 2):
fact := factor(f(Z)):
F := z -> subs(Z=z,fact):
`f(z) ` = f(z);
`Simplify the function.`;
`F(z) ` = F(z); ` `;
`Evaluate F(z) at  z = 1+i`;
`F(1+i) ` = F(1+I);

`f(z) ` = (z^2-2*I)/(z^2-2*z+2)

`Simplify the function.`

`F(z) ` = (z+(1+I))/(z+(-1+I))

` `

`Evaluate F(z) at z = 1+i`

`F(1+i) ` = 1-I

>

End of Section 2.4.