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Introduction to complex analysis
Introduction to complex analysis

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2.3 Further exercises

Here are some further exercises to end this section.

Exercise 19

Calculate the partial derivatives prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis of each of the following functions.

  • a.u of x comma y equals sum with 3 summands three times x plus x times y plus two times x squared times y squared

  • b.u of x comma y equals x times cosine of y plus exp of x times y

  • c.u of x comma y equals left parenthesis x plus y right parenthesis cubed

Answer

  • a.Differentiating u of x comma y equals sum with 3 summands three times x plus x times y plus two times x squared times y squared with respect to x while keeping y fixed, we obtain

    prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals sum with 3 summands three plus y plus four times x times y squared full stop

    Differentiating with respect to y while keeping x fixed, we obtain

    prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals x plus four times x squared times y full stop
  • b.Here u of x comma y equals x times cosine of y plus exp of x times y, so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals cosine of y plus y times exp of x times y and row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative x times sine of y plus x times exp of x times y full stop
  • c.Here u of x comma y equals left parenthesis x plus y right parenthesis cubed, so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals three times left parenthesis x plus y right parenthesis squared and row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals three times left parenthesis x plus y right parenthesis squared full stop

Exercise 20

Calculate the partial derivatives prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis of each of the following functions, and evaluate these partial derivatives at left parenthesis one comma zero right parenthesis.

  • a.u of x comma y equals x cubed times y minus y times cosine of y

  • b.u of x comma y equals y times e super x minus x times y cubed

Answer

  • a.Here u of x comma y equals x cubed times y minus y times cosine of y, so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals three times x squared times y and row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals x cubed minus cosine of y plus y times sine of y full stop

    So, at left parenthesis x comma y right parenthesis equals left parenthesis one comma zero right parenthesis the partial derivatives have the values

    prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis one comma zero right parenthesis equals zero and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis one comma zero right parenthesis equals zero full stop
  • b.Here u of x comma y equals y times e super x minus x times y cubed, so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals y times e super x minus y cubed and row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals e super x minus three times x times y squared full stop

    So, at left parenthesis x comma y right parenthesis equals left parenthesis one comma zero right parenthesis the partial derivatives have the values

    prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis one comma zero right parenthesis equals zero and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis one comma zero right parenthesis equals e full stop

Exercise 21

Find the gradient of the graph of u of x comma y equals x squared plus two times x times y at the point left parenthesis one comma two comma five right parenthesis in the x-direction and in the y-direction.

Answer

Since u of x comma y equals x squared plus two times x times y, it follows that

prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times x plus two times y and prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals two times x full stop

The gradient of the graph at left parenthesis x comma y right parenthesis equals left parenthesis one comma two right parenthesis in the x-direction is

equation sequence part 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis one comma two right parenthesis equals part 2 two multiplication one plus two multiplication two equals part 3 six full stop

The gradient of the graph at left parenthesis x comma y right parenthesis equals left parenthesis one comma two right parenthesis in the y-direction is

equation sequence part 1 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis one comma two right parenthesis equals part 2 two multiplication one equals part 3 two full stop

Exercise 22

Use the Cauchy–Riemann equations to show that there is no point of double-struck cap c at which the function

f times left parenthesis x plus i times y right parenthesis equals e super x times left parenthesis sine of y plus i times cosine of y right parenthesis

is differentiable.

Answer

Writing f in the form

f times left parenthesis x plus i times y right parenthesis equals u of x comma y plus i times v of x comma y comma

we obtain

u of x comma y equals e super x times sine of y and v of x comma y equals e super x times cosine of y full stop

Hence

multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals e super x times sine of y comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals e super x times cosine of y comma row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals e super x times cosine of y comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative e super x times sine of y full stop

If f is differentiable at x plus i times y, then the Cauchy–Riemann equations require that

multiline equation row 1 e super x times sine of y equals negative e super x times sine of y and row 2 e super x times cosine of y equals negative e super x times cosine of y semicolon

that is,

e super x times sine of y equals zero and e super x times cosine of y equals zero full stop

But e super x is never zero, so equation sequence part 1 sine of y equals part 2 cosine of y equals part 3 zero, which is impossible. It follows that there is no point of double-struck cap c at which f is differentiable.

Exercise 23

Use the Cauchy–Riemann equations to show that the function

f times left parenthesis x plus i times y right parenthesis equals left parenthesis x squared plus x minus y squared right parenthesis plus i times left parenthesis two times x times y plus y right parenthesis

is entire, and find its derivative.

Answer

In this case,

multiline equation row 1 u of x comma y equals x squared plus x minus y squared and row 2 v of x comma y equals two times x times y plus y comma

so

multiline equation row 1 Blank prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times x plus one comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times y comma row 2 Blank prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative two times y comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals two times x plus one full stop

These partial derivatives are defined and continuous on the whole of double-struck cap c. Furthermore,

multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis and row 2 prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals negative prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis comma

so the Cauchy–Riemann equations are satisfied at every point of double-struck cap c.

By the Cauchy–Riemann Converse Theorem, f is entire, and

multiline equation row 1 f super prime times left parenthesis x plus i times y right parenthesis equals prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis plus i times prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis row 2 Blank equals left parenthesis two times x plus one right parenthesis plus two times y times i full stop

(So f super prime of z equals two times z plus one, and in fact f of z equals z squared plus z.)

Exercise 24

Use the Cauchy–Riemann equations to find all the points at which the following functions are differentiable, and calculate their derivatives.

  • a.f times left parenthesis x plus i times y right parenthesis equals left parenthesis x squared plus y squared right parenthesis plus i times left parenthesis x squared minus y squared right parenthesis

  • b.f times left parenthesis x plus i times y right parenthesis equals x times y

Answer

  • a.Here

    u of x comma y equals x squared plus y squared and v of x comma y equals x squared minus y squared comma

    so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times x comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals two times x comma row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals two times y comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative two times y full stop

    The Cauchy–Riemann equations are satisfied only if x equals negative y. So f cannot be differentiable at x plus i times y unless x equals negative y. Since the partial derivatives above exist, and are continuous on double-struck cap c (and in particular when x equals negative y), it follows from the Cauchy–Riemann Converse Theorem that f is differentiable on the set x plus i times y colon x equals negative y.

    On this set,

    multiline equation row 1 f super prime times left parenthesis x plus i times y right parenthesis equals prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis plus i times prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis row 2 Blank equation sequence part 1 equals part 2 two times x plus two times x times i equals part 3 two times x times left parenthesis one plus i right parenthesis full stop
  • b.Here

    u of x comma y equals x times y and v of x comma y equals zero comma

    so

    multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals y comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals zero comma row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals x comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals zero full stop

    The Cauchy–Riemann equations are not satisfied unless y equals zero and negative x equals zero. So f is not differentiable except possibly at 0. Since the partial derivatives above exist, and are continuous at left parenthesis zero comma zero right parenthesis, it follows from the Cauchy–Riemann Converse Theorem that f is differentiable at 0. Furthermore,

    equation sequence part 1 f super prime of zero equals part 2 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis zero comma zero right parenthesis plus i times prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis zero comma zero right parenthesis equals part 3 zero full stop