Hints and Remarks for Section 2.2 (page 63) Homework Problems
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Below the following notation is used:  

"\leq"  means "less than or equal to," 
"\geq" means "greater than or equal to"
"\neq"  means  "not equal to,"
"\infty"  means  "infinity"
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1. (#15)

The hint in the book is not the right one.  Use the following full
version of the triangle inequality.

     (|a|-|b|)  \leq  (| |a|-|b| |)  \leq (|a-b|)  \leq (|a|+|b|)
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2. (#17)

(i) First let Dz=Dx.  Then,  lim as Dz -> 0 of f will be 2i.

(ii) Now let Dz=i Dy.  Then, lim of f as Dz->0 will be 1+2i.

Since the two limits are not equal, the limit does not exist.
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3. (#18)






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4. (#20)

f(z) = e^z  =>  u(x,y) = Re(f) = e^x cos y
		v(x,y) = Im(f) = e^x sin y
Now, from real analysis we know that u and v are continuous functions
(actually they are even C^(\infty)).  Therefore, we can write

 limit of u(x,y) as (x,y) -> (x_0,y_0) = u(x_0,y_0) and similarly for
 v(x,y).  Hence, using the results of problem #18 we get the desired
 results.
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5. (#22)

The proof is by contradiction.  So,

(i)  assume that lim as n-> \infty f(z_n) = w_0 for ALL sequences
     z_n -> z_0.

(ii) assume that lim as z -> z_0 of f(z) \neq w_0.

(iii) Now we want to arrive at a contradiction.  What we need is one
      sequence z_n -> z_0 along which f(z_n) does not converge to w_0.

But the second assumption implies that

  there exists an epsilon >0 for which for ALL delta |z-z_0| < delta
  implies that |f(z)-w_0| > epsilon.  Hence, if for THIS epsilon we
  let z_n=z_0+(1/n) we get the desired result.  You can complete the
  rest of the argument.
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