1.6 Further exercises

Here are some further exercises to end this section.

Exercise 11

Use the definition of derivative to find the derivative of the function

f of z equals two times z squared plus five full stop

Answer

The function f of z equals two times z squared plus five is defined on the whole of double-struck cap c. Let alpha element of double-struck cap c. Then

multiline equation row 1 f super prime of alpha equals lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha row 2 Blank equals lim over z right arrow alpha of left parenthesis two times z squared plus five right parenthesis minus left parenthesis two times alpha squared plus five right parenthesis divided by z minus alpha row 3 Blank equals lim over z right arrow alpha of two times left parenthesis z squared minus alpha squared right parenthesis divided by z minus alpha row 4 Blank equals lim over z right arrow alpha of two times left parenthesis z plus alpha right parenthesis row 5 Blank equals four times alpha full stop

Since alpha is an arbitrary complex number, f is differentiable on the whole of double-struck cap c, and the derivative is the function

f super prime of z equals four times z times left parenthesis z element of double-struck cap c right parenthesis full stop

Exercise 12

Prove the Quotient Rule for differentiation.

Answer

Let cap f equals f solidus g. Then

multiline equation row 1 Blank cap f of z minus cap f of alpha divided by z minus alpha Blank row 2 Blank equals f of z solidus g of z minus f of alpha solidus g of alpha divided by z minus alpha row 3 Blank equals f of z times g of alpha minus f of alpha times g of z divided by left parenthesis z minus alpha right parenthesis times g of z times g of alpha Blank row 4 Blank equals g of alpha times left parenthesis f of z minus f of alpha right parenthesis minus f of alpha times left parenthesis g of z minus g of alpha right parenthesis divided by left parenthesis z minus alpha right parenthesis times g of z times g of alpha Blank row 5 Blank equals g of alpha times left parenthesis f of z minus f of alpha divided by z minus alpha right parenthesis minus f of alpha times left parenthesis g of z minus g of alpha divided by z minus alpha right parenthesis divided by g of z times g of alpha full stop Blank

Using the Combination Rules for limits of functions, the continuity of g, and the fact that g of alpha not equals zero, we can take limits to obtain

cap f super prime of alpha equals g of alpha times f super prime of alpha minus f of alpha times g super prime of alpha divided by left parenthesis g of alpha right parenthesis squared full stop

Exercise 13

Find the derivative of each of the following functions f. In each case specify the domain of f super prime.

  1. f of z equals sum with 3 summands z squared plus two times z plus one divided by three times z plus one

  2. f of z equals z cubed plus one divided by z squared minus z minus six

  3. f of z equals one divided by sum with 3 summands z squared plus two times z plus two

  4. f of z equals sum with 3 summands z squared plus five times z minus two plus one divided by z plus one divided by z squared

Answer

  1. By the Combination Rules,

    multiline equation row 1 f super prime of z equals left parenthesis three times z plus one right parenthesis times left parenthesis two times z plus two right parenthesis minus three times left parenthesis sum with 3 summands z squared plus two times z plus one right parenthesis divided by left parenthesis three times z plus one right parenthesis squared row 2 Blank equals three times z squared plus two times z minus one divided by left parenthesis three times z plus one right parenthesis squared full stop

    The domain of f super prime is double-struck cap c minus left curly bracket negative one solidus three right curly bracket.

  2. By the Combination Rules,

    multiline equation row 1 f super prime of z equals left parenthesis z squared minus z minus six right parenthesis times left parenthesis three times z squared right parenthesis minus left parenthesis z cubed plus one right parenthesis times left parenthesis two times z minus one right parenthesis divided by left parenthesis z squared minus z minus six right parenthesis squared Blank row 2 Blank equals z super four minus two times z cubed minus 18 times z squared minus two times z plus one divided by left parenthesis z squared minus z minus six right parenthesis squared full stop Blank

    Since z squared minus z minus six equals left parenthesis z plus two right parenthesis times left parenthesis z minus three right parenthesis, the domain of f super prime is double-struck cap c minus negative two comma three.

  3. By the Reciprocal Rule,

    f super prime of z equals negative left parenthesis two times z plus two right parenthesis divided by left parenthesis sum with 3 summands z squared plus two times z plus two right parenthesis squared full stop

    The roots of sum with 3 summands z squared plus two times z plus two are negative two plus minus Square root of negative four divided by two equals negative one plus minus i. The domain of f super prime is therefore double-struck cap c minus negative one plus i comma negative one minus i.

  4. By the Sum Rule and the rule for differentiating integer powers,

    f super prime of z equals two times z plus five minus one divided by z squared minus two divided by z cubed full stop

    The domain of f super prime is double-struck cap c minus zero.

Exercise 14

Use Strategy B to show that there are no points of double-struck cap c at which the function

f of z equals Im of z

is differentiable.

Answer

Consider an arbitrary complex number alpha equals a plus i times b, where a comma b element of double-struck cap r. Let z sub n equals alpha plus one solidus n, n equals one comma two comma ellipsis. Then z sub n right arrow alpha, and

multiline equation row 1 lim over n right arrow normal infinity of Im of z sub n minus Im of alpha divided by z sub n minus alpha equals lim over n right arrow normal infinity of b minus b divided by one solidus n row 2 Blank equation sequence part 1 equals part 2 lim over n right arrow normal infinity of zero divided by one solidus n equals part 3 zero full stop

Now let z sub n super prime equals alpha plus i solidus n, n equals one comma two comma ellipsis. Then z sub n super prime right arrow alpha, and

multiline equation row 1 lim over n right arrow normal infinity of Im of z sub n super prime minus Im of alpha divided by z sub n super prime minus alpha equals lim over n right arrow normal infinity of left parenthesis b plus one solidus n right parenthesis minus b divided by i solidus n row 2 Blank equation sequence part 1 equals part 2 lim over n right arrow normal infinity of one solidus n divided by i solidus n equals part 3 negative i full stop

Since the two limits do not agree, it follows that Im fails to be differentiable at each point of double-struck cap c.

Exercise 15

Describe the approximate geometric effect of the function

f of z equals z cubed plus eight divided by z minus six

on a small disc centred at the point 2.

Answer

To a close approximation, a small disc centred at 2 is mapped by f to small disc centred at f of two equals negative four. In the process, the disc is scaled by the factor absolute value of f super prime of two and rotated through the angle Arg of f super prime of two.

By the Quotient Rule,

multiline equation row 1 f super prime of z equals three times z squared times left parenthesis z minus six right parenthesis minus left parenthesis z cubed plus eight right parenthesis divided by left parenthesis z minus six right parenthesis squared row 2 Blank equals two times z cubed minus 18 times z squared minus eight divided by left parenthesis z minus six right parenthesis squared full stop

So f scales the disc by the factor 4 and rotates it anticlockwise through the angle pi.