Readiness

Practice: Chapter review: complex arithmetic (4 problems · doesn’t affect your progress)

Compute $(6-i)+(-2+5i)$ . Enter your answer as a complex number.

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Why

$(6-2)+(-1+5)i=4+4i$ .

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Why

4 + 4*I

Find the product $(2+3i)(4-i)$ , writing the result as a single complex number.

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Why

$(2+3i)(4-i)=8-2i+12i-3i^2=8+10i+3=11+10i$ .

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Why

Expand like two binomials, then simplify $i^2$ .

11 + 10*I

Find $|z|$ for $z=8+6i$ .

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Why

$|z|=\sqrt{64+36}=\sqrt{100}=10$ .

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Why

10

What is $\overline{4-7i}$ ?

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The real part, $4$ , stays exactly as it was — only the imaginary part's sign flips.

This flips both signs, giving $-z$ instead of $\bar z$ .

Swapping the two parts is a different operation entirely, not conjugation.

Why

Only the sign in front of $i$ flips: $\overline{4-7i}=4+7i$ .

$4+7i$

Practice: Chapter review: polar and exponential form (4 problems · doesn’t affect your progress)

Find $r$ for $z=6i$ .

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Why

$r=\sqrt{0^2+6^2}=6$ .

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6

Find $\operatorname{Arg}(z)$ for $z=6i$ , in terms of $\pi$ .

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Why

$6i$ points straight up the imaginary axis, angle $\pi/2$ .

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Why

pi/2

Evaluate $e^{i2\pi/3}$ in $a+bi$ form.

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Why

$e^{i2\pi/3}=\cos(2\pi/3)+i\sin(2\pi/3)=-\tfrac12+\tfrac{\sqrt3}{2}i$ .

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Why

$\cos(2\pi/3)=-1/2$ , $\sin(2\pi/3)=\sqrt3/2$ .

-1/2 + sqrt(3)/2*I

Which of these has the same value as $4e^{i\pi}$ ?

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$e^{i\pi}=\cos\pi+i\sin\pi=-1$ , not $1$ — the angle $\pi$ points backward, not forward.

That would require the angle to be $\pi/2$ , not $\pi$ .

That would require the angle to be $-\pi/2$ or $3\pi/2$ , not $\pi$ .

Why

$4e^{i\pi}=4(\cos\pi+i\sin\pi)=4(-1+0i)=-4$ .

$-4$

Practice: Chapter review: De Moivre's theorem and roots (4 problems · doesn’t affect your progress)

Find $(1+i\sqrt3)^3$ using De Moivre's theorem.

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Why

$(2e^{i\pi/3})^3=8e^{i\pi}=8(-1+0i)=-8$ .

Correct answer

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Why

$1+i\sqrt3=2e^{i\pi/3}$ .

-8

Every nonzero complex number has exactly how many distinct 8th roots?

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Why

An $n$ th root equation always has exactly $n$ distinct solutions for a nonzero right-hand side.

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8

Find the 4th root of unity with positive imaginary part.

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Why

The root at angle $\pi/2$ is $\cos(\pi/2)+i\sin(\pi/2)=i$ .

Correct answer

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Why

The 4th roots of unity sit at angles $0,\pi/2,\pi,3\pi/2$ .

I

A cube root of a complex number $w$ has modulus $2$ . What is $|w|$ ?

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Why

If $|w|^{1/3}=2$ , then $|w|=2^3=8$ .

Correct answer

Also accepted:

Why

An $n$ th root's modulus is $|w|^{1/n}$ .

8

Practice: Chapter review: the Fundamental Theorem of Algebra (4 problems · doesn’t affect your progress)

How many roots (with multiplicity) does a degree-4 polynomial have over $\mathbb C$ ?

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Why

The theorem guarantees exactly $4$ roots, counting multiplicity.

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4

A real cubic has roots $2$ , $1+i$ , and its guaranteed conjugate partner. State the third root.

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Why

Nonreal roots of a real polynomial come with their conjugate: $\overline{1+i}=1-i$ .

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1 - I

A real-coefficient polynomial has a nonreal root with multiplicity $3$ . What else must be true?

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Repeated nonreal roots are perfectly possible — multiplicity applies to any root, real or not.

Multiplicities of a conjugate pair have to match exactly — one at $3$ and the other at $1$ would break the pairing.

Real coefficients are exactly what forces the pairing in the first place — this scenario is fully consistent with real coefficients.

Why

Conjugating $p(z)=0$ shows $\bar z$ is a root of the same multiplicity as $z$ whenever the coefficients are real.

Its conjugate must also be a root, with the same multiplicity, $3$

A real-coefficient degree-8 polynomial has exactly $0$ real roots. How many conjugate pairs does it have?

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Why

$8$ nonreal roots split into $8/2=4$ conjugate pairs.

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Why

All $8$ roots are nonreal, split evenly into pairs.

4

Practice: Chapter review: sequences and series (4 problems · doesn’t affect your progress)

Find $\displaystyle\lim_{n\to\infty}\frac{4n+7}{2n-1}$ .

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Why

Dividing by $n$ : $\dfrac{4+7/n}{2-1/n}\to2$ .

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Why

2

Find the sum of $\displaystyle\sum_{n=0}^\infty 6\left(\frac13\right)^n$ .

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Why

$\dfrac{6}{1-1/3}=\dfrac{6}{2/3}=9$ .

Correct answer

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Why

Use $\dfrac{a}{1-r}$ with $a=6$ , $r=1/3$ .

9

Apply the ratio test to $\displaystyle\sum \frac{n^2}{5^n}$ . Report the limit $L$ as a fraction.

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Why

$\left(\dfrac{n+1}{n}\right)^2\cdot\dfrac15\to1\cdot\dfrac15=\dfrac15$ .

Correct answer

Also accepted:

Why

The $\left(\frac{n+1}{n}\right)^2$ factor goes to $1$ as $n\to\infty$ .

1/5

Does $\sum(3/4)^n$ converge?

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$|r|=3/4$ is below $1$ , which is exactly the convergence condition.

The ratio $r=3/4$ is all that's needed to decide this — it's already fully determined.

Where the sum starts doesn't change whether it converges — only $r$ does.

Why

A geometric series converges exactly when $|r|<1$ ; here $|r|=3/4<1$ .

Yes — geometric with $|r|=3/4<1$

Practice: Chapter review: power series and radius of convergence (4 problems · doesn’t affect your progress)

What is the radius of convergence of $\displaystyle\sum \frac{x^n}{7^n}$ ?

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$\lim|x/7|<1 \Leftrightarrow |x|<7$ .

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7

Find the radius of convergence of $\displaystyle\sum \frac{(x-2)^n}{n\cdot5^n}$ , centered at $2$ .

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Why

$\lim\dfrac15|x-2|<1 \Leftrightarrow |x-2|<5$ .

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5

What is the radius of convergence of the Maclaurin series for $e^x$ ?

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$R=1$ would fail for $x=2$ , but $e^2$ 's series converges perfectly well, just like every other real number.

$e$ has no special role in where this particular series converges.

$R=0$ would mean the series only converges at $x=0$ — but the ratio test's limit is $0$ for every $x$ .

Why

The ratio test's limit is $0$ for every real $x$ , since $n!$ grows without bound — the series converges everywhere.

$\infty$

A series has $R=6$ centered at $c=0$ . Is convergence at $x=6$ guaranteed?

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The ratio test settles $|x|<R$ versus $|x|>R$ only — it says nothing about $|x|=R$ exactly.

Plenty of series do converge at an endpoint — it's just never automatic.

Being geometric doesn't change this rule — even a geometric series' endpoint behavior follows from its own separate check ( $|r|=1$ there), not from $R$ .

Why

The endpoint $x=R$ (and $x=-R$ ) always needs a separate convergence test — the ratio test itself is silent about exactly $|x-c|=R$ .

No — the endpoint needs separate testing

Practice: Chapter review: partial derivatives, the Jacobian, and Green's theorem (4 problems · doesn’t affect your progress)

For $f(x,y)=5xy^2$ , find $f_x$ .

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Why

Freezing $y$ , the derivative of $5xy^2$ in $x$ is $5y^2$ .

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Why

5*y^2

For the same $f(x,y)=5xy^2$ , now find the partial derivative with respect to $y$ , $f_y$ .

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Why

Freezing $x$ , the derivative of $5xy^2$ in $y$ is $10xy$ .

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Why

10xy

For $u=2x+y$ , $v=x-3y$ , find the Jacobian determinant $\partial(u,v)/\partial(x,y)$ .

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Why

Determinant $=(2)(-3)-(1)(1)=-6-1=-7$ .

Correct answer

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Why

$u_x=2$ , $u_y=1$ , $v_x=1$ , $v_y=-3$ .

-7

Green's theorem converts a line integral around a closed curve into...

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Green's theorem stays two-dimensional throughout — no third dimension or solid ever appears.

The whole point is trading a boundary computation for an interior one — not swapping curves.

The right-hand side is a genuine double integral over a flat piece of the plane, not a single-variable one.

Why

$\oint_C P\,dx+Q\,dy = \iint (Q_x-P_y)\,dA$ over the flat area $C$ encloses.

A double integral over the flat area the curve encloses

Closing thought

Recipes in this chapter

Chapter review