Complex Analysis Worksheet 5- Real Part, Imaginary Part

Complex Analysis Worksheet 5- Real Part, Imaginary Part, Functional Limits
• Any complex function f (z) can be thought of as two functions from R2 to R, namely
Re(f (z)) and Im(f (z)).
• Example We revisit z → z 2 ,
Which takes a + bi → a2 − b2 + 2abi,
So Re(f (z)) = a2 − b2 and Im(f (z)) = 2ab which are 2 functions from R2 to R.
• Using this, we can can look at a different representation of the graph of z → z 2 by
looking at how it transforms the right half plane with horizontal and vertical lines
(they become parabolas).
• Limits of complex functions (as opposed to sequences)
Working Definition We say
lim f (z) = L
if as |z − z0 | → 0 then |f (z) − L| → 0.
Notes: (a) The above is modulus, not absolute value
(b) L above is a complex number (it could have 0 imaginary part though :) ).
(c) The limits above are actually REAL limits, so we are not being circular.
lim Argz = d.n.e.
Since there are 2 paths that disagree (recall limits from calc 3!)
Example: Use the ’working definition’ to verify that limz→−2i z 3 = 8i. (Uses complex
long division)
Example: Compute
lim √
and then use the definition to prove it is true.
Challenging Example: Compute limz→−1/2+√3/2i
definition to prove it is true.
z 3 −1√
z−(−1/2+ 3/2i)
and then use the
• The technical definition:
lim f (z) = L
For any ² > 0 there exists δ > 0 so that whenever 0 < |z − z0 | < δ then |f (z) − L| < ².
Note: We will not stress the technical definition at this point.
• Limit Theorems (not proven, proofs are almost identical to those in real analysis)
(a) limits are unique
(b) The limit of the sum is the sum of the limits.
(c) The limit of the product is the product of the limits.
(d) Same with differences and non-zero quotients.
Notes: 0/0 is still indeterminate. number/0 is still ”infinite” or at least d.n.e., also
you may use the ’plug in method’ unless I ask you to use the definition.
• Definition (Continuity) we say that f (z) is continuous at z0 if ...
HW #5:
(1) Recreate the analysis done in class by analyzing what happens to vertical and
horizontal lines in the first quadrant under the function f (z) = z 4
(2) Prove that limz→0
does not exist. (Hint, approach the origin along different
(3) Use the working definition to show that limz→i z 4 = 1
(4) Compute limz→i
z 4 −1
and use the working definition to verify your answer.