Make learning comfortable

Visual experienceAnimationRead-aloud speedChanging style, motion, or reading speed keeps your current question, answers, homework, and saved progress. The school band still chooses your task layout. Saved to your learner profile across devices. All lessons and help stay available.

← Complex Analysis: Analytic Functions levels

Complex Analysis: Analytic Functions / LEVEL 1 · DIFFICULTY 4/5

Complex Analysis: Analytic Functions

Read the prerequisite, study a worked example and justify each step in these selected university topics.

2 stages · 16 practice problems · two 4-question assessment forms

Choose a node to explore its lesson.

MINI QUEST UCX 1.1Analyticity & Cauchy–RiemannRead the lesson
MINI QUEST UCX 1.2Contour Integrals, Laurent Terms & ResiduesRead the lesson
LEVEL CHECKCastle challenge
4 questions across this level.
Sign in for the level check →

STAGE UCX 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Analyticity & Cauchy–Riemann

Useful preparation: Integrals & the Fundamental Theorem

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A complex derivative describes change from every direction using one complex number. Its real linear approximation must therefore act like complex multiplication.

Complex differentiability: For a function defined near $\displaystyle z_{0}$ , the derivative is the limit of $\displaystyle \frac{f\, \left(z_{0} + h\right) - f\, \left(z_{0}\right)}{h}$ as the nonzero complex increment h tends to zero. The limit must be independent of the approach.
Holomorphic and analytic: A function is holomorphic on an open set if it is complex differentiable at every point there. In one complex variable this is equivalent to locally having a convergent power series, called analyticity.
Cauchy–Riemann equations: Write $\displaystyle z = x$ +iy and $\displaystyle f = u$ +iv with real-valued u,v. The equations $\displaystyle u_{x} = v$ ᵧ and uᵧ= $\displaystyle -v_{x}$ connect their partial derivatives; subscripts specify the real coordinate differentiated.

Conditions before calculation

Complex differentiability at a point implies the Cauchy–Riemann equations there. Their satisfaction alone at one point is not a sufficient differentiability test.
A sufficient pointwise test is that u and v have continuous first partial derivatives in a neighborhood and satisfy the equations at the point. If they satisfy them throughout an open set, f is holomorphic there.
Differentiability at one isolated point does not imply holomorphicity on any neighborhood. When using quotient or logarithm formulas, exclude zeros or specify an appropriate branch and domain.

Why the method works

1. Compare two necessary limits

Approach with a real increment and then a purely imaginary increment. If the complex derivative exists, both resulting expressions must agree, giving the equations.

f'(z_0)=u_x+iv_x=v_y-iu_y

2. Use real differentiability for sufficiency

Continuous partials provide a real linear approximation. Its error divided by $\displaystyle \left|h\right|$ tends to zero, denoted $\displaystyle o\, \left(\left|h\right|\right)$ . The equations turn the linear part into multiplication by $\displaystyle u_{x}$ +ivₓ.

f(z_0+h)-f(z_0)=(u_x+iv_x)h+o(|h|)

3. Check the scope of the conclusion

After computing a derivative, state where it exists. Passing the test only at one point supports only a pointwise conclusion, not analyticity of a whole region.

Worked examples

EXAMPLE 1

Verify that $\displaystyle f\, \left(z\right) = z^{2} + \left(2 - i\right)\, z$ is holomorphic and compute $\displaystyle f'\, \left(1 - i\right)$ .

Expand using $\displaystyle z = x$ +iy to identify the real and imaginary parts.
$u=x^2-y^2+2x+y,\qquad v=2xy+2y-x$
The partials are continuous polynomials and satisfy both equations everywhere.
$u_x=v_y=2x+2,\qquad u_y=-2y+1=-v_x$
Use $\displaystyle u_{x}$ +ivₓ for the derivative and substitute the requested point.
$f'(z)=2z+2-i,\qquad f'(1-i)=4-3i$

Conclusion: The polynomial is holomorphic on the entire plane, and the derivative is $\displaystyle 4 - 3\, i$ at $\displaystyle 1 - i$ .

EXAMPLE 2

Determine where $\displaystyle g\, \left(z\right) = \left|z - 2\right|^{2}$ is complex differentiable.

Here $\displaystyle u = \left(x - 2\right)^{2} + y^{2}$ and $\displaystyle v = 0$ . The equations force $\displaystyle x = 2$ and $\displaystyle y = 0$ , so no other point can work.
At $\displaystyle z = 2$ , evaluate the difference quotient directly.
$\frac{g(2+h)-g(2)}h=\frac{|h|^2}h=\overline h$
Its modulus equals $\displaystyle \left|h\right|$ and tends to zero along every approach, proving $\displaystyle g'\, \left(2\right) = 0$ .

Conclusion: It is differentiable only at 2 and is holomorphic on no nonempty open set.

Pitfalls & counterexamples

Tempting claim: Cauchy–Riemann equations at one point always prove differentiability.

Correction: A regularity assumption is needed for this converse.

Test the claim: Set $\displaystyle u = \frac{xy}{\sqrt{x^{2} + y^{2}}}$ away from the origin, $\displaystyle u\, \left(0, 0\right) = 0$ , and $\displaystyle v = 0$ . All partials at zero vanish, but the quotient is zero along real increments and $\displaystyle \frac{1}{\sqrt{2}\, \left(1 + i\right)}$ along $\displaystyle h = t\, \left(1 + i\right)$ , $\displaystyle t > 0$ .

Check your understanding

Is the conjugation map $\displaystyle f\, \left(z\right) = z$ ̄ complex differentiable anywhere?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

Its difference quotient is 1 for nonzero real increments and −1 for purely imaginary increments.

Connect this idea

The equations connect complex differentiation with real Jacobian matrices. Holomorphicity on regions is the foundation for Cauchy’s theorem, local power series, and residue calculations.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

MIT OCW 18.04, Topic 2: Analytic functions

Preview the eight practice prompts

Problem 1
For $\displaystyle f\, \left(z\right) = z^{2} + 2$ , find the real part of $\displaystyle f'\, \left(1 + i\right)$ .
Practise this exact problem →
Problem 2
For $\displaystyle f\, \left(z\right) = 2\, z^{3}$ , find $\displaystyle f'\, \left(1\right)$ .
Practise this exact problem →
Problem 3
Let $\displaystyle u\, \left(x, y\right) = 2\, x$ and $\displaystyle v\, \left(x, y\right) = -2\, y$ . Do the Cauchy–Riemann equations hold? Enter 1 for yes, 0 for no.
Practise this exact problem →
Problem 4
For $\displaystyle u = x^{2} - y^{2}$ and $\displaystyle v = a\, xy$ , determine a so the Cauchy–Riemann equations hold near (1,1).
Practise this exact problem →
Problem 5New context
A complex model uses $\displaystyle f\, \left(z\right) = z^{2} + 3$ , find the real part of $\displaystyle f'\, \left(1 + i\right)$ .
Practise this exact problem →
Problem 6New context
A complex field is $\displaystyle f\, \left(z\right) = 3\, z^{3}$ , find $\displaystyle f'\, \left(1\right)$ .
Practise this exact problem →
Problem 7New context
For $\displaystyle f = u$ +iv, let $\displaystyle u\, \left(x, y\right) = 3\, x$ and $\displaystyle v\, \left(x, y\right) = -3\, y$ . Do the Cauchy–Riemann equations hold? Enter 1 for yes, 0 for no.
Practise this exact problem →
Problem 8New context
A candidate analytic function has $\displaystyle u = x^{2} - y^{2}$ and $\displaystyle v = a\, xy$ , determine a so the Cauchy–Riemann equations hold near (2,1).
Practise this exact problem →

Open stage UCX 1.1 in the student workspace →

STAGE UCX 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Contour Integrals, Laurent Terms & Residues

Useful preparation: Analyticity & Cauchy–Riemann

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Residues are the reciprocal-term coefficients that survive integration around singularities. The residue theorem combines these local contributions; contour orientation determines their signs.

Contour integral: For a piecewise continuously differentiable path $\displaystyle \gamma : \left[a, b\right] \to C$ and a continuous integrand along it, integrate $\displaystyle f\, \left(\gamma\, \left(t\right)\right)\, \gamma'\, \left(t\right)$ over the parameter interval. Reversing the path reverses the sign.
Laurent series and residue: A Laurent expansion around a includes integer powers, possibly negative, of $\displaystyle z - a$ and converges on an annulus. Its coefficient of $\displaystyle \left(z - a\right)^{-1}$ is the residue at an isolated singularity a.
Pole: A pole of order m has a finite negative-power part with highest reciprocal power $\displaystyle \left(z - a\right)^{-m}$ and a nonzero coefficient there. A simple pole has $\displaystyle m = 1$ ; a removable singularity is not a pole.

Conditions before calculation

For the version used here, C is a piecewise smooth simple closed contour, traversed once counterclockwise. The function is holomorphic on an open neighborhood of the contour and its interior except at finitely many isolated poles inside.
No pole may lie on C. Only enclosed poles contribute. Clockwise traversal multiplies the result by −1; more general closed curves require winding numbers rather than a simple inside/outside count.
The simple-pole formula $\displaystyle \frac{p\, \left(a\right)}{q'}\, \left(a\right)$ applies when p,q are holomorphic near a, $\displaystyle q\, \left(a\right) = 0$ , $\displaystyle q'\, \left(a\right) \ne 0$ , and $\displaystyle p\, \left(a\right) \ne 0$ . If the numerator also vanishes, examine cancellation first.

Why the method works

1. Locate singularities before computing

Factor the denominator and check numerator cancellation. Compare each remaining pole with the actual contour; a correct residue at an exterior pole contributes nothing.

2. Extract the reciprocal coefficient

For a simple pole, multiplying by $\displaystyle z - a$ removes its singular factor. Evaluating the remaining holomorphic expression gives the residue.

\operatorname{Res}(f,a)=\lim_{z\to a}(z-a)f(z)

3. Account for the orientation

One counterclockwise turn contributes $\displaystyle 2\, \pi\, i$ times each enclosed residue. This comes from integrating the reciprocal term; other Laurent powers integrate to zero around a small circle.

\oint_C f(z)\,dz=2\pi i\sum_{a\text{ inside }C}\operatorname{Res}(f,a)

Worked examples

EXAMPLE 1

Integrate $\displaystyle f\, \left(z\right) = \frac{z + 1}{\left(z - 2\right)\, \left(z + 3\right)}$ around $\displaystyle \left|z\right| = \frac{5}{2}$ counterclockwise.

The simple poles are 2 and −3. Neither is on the circle; only 2 has modulus less than $\displaystyle \frac{5}{2}$ .
Remove the factor $\displaystyle z - 2$ and substitute 2.
$\operatorname{Res}(f,2)=\left.\frac{z+1}{z+3}\right|_{z=2}=\frac35$
Multiply the single enclosed residue by $\displaystyle 2\, \pi\, i$ . The exterior pole is excluded.
$\oint_C f(z)\,dz=\frac{6\pi i}{5}$

Conclusion: The integral is $\displaystyle \frac{6\, \pi\, i}{5}$ ; clockwise traversal would give $\displaystyle \frac{-6\, \pi\, i}{5}$ .

EXAMPLE 2

Integrate $\displaystyle g\, \left(z\right) = \frac{z^{2} + 1}{\left(z - 1\right)\, \left(z + 2\right)}$ around $\displaystyle \left|z\right| = 3$ clockwise.

Both simple poles, 1 and −2, lie strictly inside the circle, and there are no poles on it.
Compute each residue after canceling its own simple-pole factor.
$\operatorname{Res}(g,1)=\frac23,\qquad\operatorname{Res}(g,-2)=-\frac53$
Their sum is −1. Clockwise orientation uses $\displaystyle -2\, \pi\, i$ times that sum.
$\oint_C g(z)\,dz=-2\pi i(-1)=2\pi i$

Conclusion: The integral is $\displaystyle 2\, \pi\, i$ ; reversing orientation changes its sign.

Pitfalls & counterexamples

Tempting claim: Any denominator zero on or inside the contour can simply be added to the residue sum.

Correction: Poles on the contour invalidate the stated theorem and may make the ordinary contour integral undefined.

Test the claim: For $\displaystyle f\, \left(z\right) = \frac{1}{z - 1}$ on the unit circle, the path passes through $\displaystyle z = 1$ , where the integrand is undefined.

Check your understanding

What is the residue at 0 of $\displaystyle \frac{4}{z^{3}} - \frac{7}{z} + 2$ , and its integral around a counterclockwise unit circle?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

The residue is −7 and the integral is $\displaystyle -14\, \pi\, i$ .

Only the coefficient of $\displaystyle z^{-1}$ contributes, although the singularity is a pole of order 3.

Connect this idea

Laurent expansions separate local singular behavior. Residues combine that local information with contour geometry and orientation to evaluate otherwise difficult integrals.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts

Problem 1
Integrate $\displaystyle \frac{2}{z}$ counterclockwise once around $\displaystyle \left|z\right| = 1$ . Write the integral as $\displaystyle a \cdot 2\, \pi\, i$ . Find a.
Practise this exact problem →
Problem 2
Use Cauchy’s formula $\displaystyle \frac{z + 2}{z - 1}$ counterclockwise once around $\displaystyle \left|z\right| = 2$ . Write the integral as $\displaystyle a \cdot 2\, \pi\, i$ . Find a.
Practise this exact problem →
Problem 3
A Laurent expansion $\displaystyle \frac{2}{z^{2}} + \frac{5}{z} + 7 + z$ . Find its residue at zero.
Practise this exact problem →
Problem 4
Integrate $\displaystyle \frac{3}{z}$ clockwise once around $\displaystyle \left|z\right| = 1$ . Write the integral as $\displaystyle a \cdot 2\, \pi\, i$ . Find a.
Practise this exact problem →
Problem 5New context
A contour calculation integrates $\displaystyle \frac{3}{z}$ counterclockwise once around $\displaystyle \left|z\right| = 1$ . Write the integral as $\displaystyle a \cdot 2\, \pi\, i$ . Find a.
Practise this exact problem →
Problem 6New context
A contour integral uses $\displaystyle \frac{z + 3}{z - 1}$ counterclockwise once around $\displaystyle \left|z\right| = 2$ . Write the integral as $\displaystyle a \cdot 2\, \pi\, i$ . Find a.
Practise this exact problem →
Problem 7New context
Near zero, a meromorphic function has expansion $\displaystyle \frac{3}{z^{2}} + \frac{6}{z} + 7 + z$ . Find its residue at zero.
Practise this exact problem →
Problem 8New context
An oriented contour follows $\displaystyle \frac{4}{z}$ clockwise once around $\displaystyle \left|z\right| = 1$ . Write the integral as $\displaystyle a \cdot 2\, \pi\, i$ . Find a.
Practise this exact problem →

Open stage UCX 1.2 in the student workspace →

Browse another level →

Complex Analysis: Analytic Functions

Analyticity & Cauchy–Riemann

Meaning & notation

Conditions before calculation

Why the method works

1. Compare two necessary limits

2. Use real differentiability for sufficiency

3. Check the scope of the conclusion

Worked examples

Verify that f (z)=z2+(2−i) z\displaystyle f\, \left(z\right) = z^{2} + \left(2 - i\right)\, zf(z)=z2+(2−i)z is holomorphic and compute f′ (1−i)\displaystyle f'\, \left(1 - i\right)f′(1−i).

Determine where g (z)=∣z−2∣2\displaystyle g\, \left(z\right) = \left|z - 2\right|^{2}g(z)=∣z−2∣2 is complex differentiable.

Pitfalls & counterexamples

Check your understanding

Connect this idea

Reference reading

Contour Integrals, Laurent Terms & Residues

Meaning & notation

Conditions before calculation

Why the method works

1. Locate singularities before computing

2. Extract the reciprocal coefficient

3. Account for the orientation

Worked examples

Integrate f (z)=z+1(z−2) (z+3)\displaystyle f\, \left(z\right) = \frac{z + 1}{\left(z - 2\right)\, \left(z + 3\right)}f(z)=(z−2)(z+3)z+1​ around ∣z∣=52\displaystyle \left|z\right| = \frac{5}{2}∣z∣=25​ counterclockwise.

Integrate g (z)=z2+1(z−1) (z+2)\displaystyle g\, \left(z\right) = \frac{z^{2} + 1}{\left(z - 1\right)\, \left(z + 2\right)}g(z)=(z−1)(z+2)z2+1​ around ∣z∣=3\displaystyle \left|z\right| = 3∣z∣=3 clockwise.

Pitfalls & counterexamples

Check your understanding

Connect this idea

Reference reading

Verify that $\displaystyle f\, \left(z\right) = z^{2} + \left(2 - i\right)\, z$ is holomorphic and compute $\displaystyle f'\, \left(1 - i\right)$ .

Determine where $\displaystyle g\, \left(z\right) = \left|z - 2\right|^{2}$ is complex differentiable.

Integrate $\displaystyle f\, \left(z\right) = \frac{z + 1}{\left(z - 2\right)\, \left(z + 3\right)}$ around $\displaystyle \left|z\right| = \frac{5}{2}$ counterclockwise.

Integrate $\displaystyle g\, \left(z\right) = \frac{z^{2} + 1}{\left(z - 1\right)\, \left(z + 2\right)}$ around $\displaystyle \left|z\right| = 3$ clockwise.