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Complex Numbers: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 A distance hidden in an equation

Difficulty: 1 of 3 stars, Stretch

A complex number $z$ satisfies $z\overline z+z+\overline z=8$ , where $\overline z$ is its complex conjugate. Find the least and greatest possible values of $|z-(2+4i)|$ , and identify every $z$ attaining either value.

Builds on Completing the Square
1. Hint 1
  Write $z=x+yi$ and complete a square. What is the set of allowed points in the complex plane?
2. Hint 2
  The allowed points lie on a circle centered at $-1$ . Compare the distance from that center to $2+4i$ with the radius.
Answer

The minimum is $2$ , attained only at $z=(4+12i)/5$ . The maximum is $8$ , attained only at $z=(-14-12i)/5$ .

Full solution

For $z=x+yi$ , the constraint is $x^2+y^2+2x=8$ , or
$(x+1)^2+y^2=9.$
Thus the allowed points form the circle of radius $3$ centered at $C=-1$ .
The target point is $A=2+4i$ , and
$|A-C|=|3+4i|=5.$
For any allowed point $Z$ , the triangle inequality gives
$AZ\le AC+CZ=8.$
Applying it instead to $AC$ gives $AC\le AZ+ZC$ , so $AZ\ge2$ .

The lower equality requires $Z$ on the segment from $C$ to $A$ , at distance $3$ from $C$ ; this gives
$z=-1+(3/5)(3+4i)=(4+12i)/5.$
The upper equality requires $C$ between $A$ and $Z$ , giving
$z=-1-(3/5)(3+4i)=(-14-12i)/5.$
These are the unique points on those respective rays at radius $3$ .
Both satisfy the circle equation, and their distances to $A$ are exactly $2$ and $8$ .
The geometric bounds therefore give all equality cases as well as both extrema.

Answer

The minimum is $2$ , attained only at $z=(4+12i)/5$ . The maximum is $8$ , attained only at $z=(-14-12i)/5$ .

Key idea

An equation involving a number and its conjugate may encode a familiar locus; distance bounds then become geometric.
Problem 1: Mark complete
Problem 2 Four applications of one rule

Difficulty: 1 of 3 stars, Stretch

Define $T(z)=(1+i)z+2-i$ . Find all complex numbers $z$ such that applying $T$ four times gives $\overline z$ . Find a useful center for the transformation so that you do not have to expand four nested expressions.
1. Hint 1
  Look for a point $c$ with $T(c)=c$ . Subtracting this point from both input and output removes the added constant.
2. Hint 2
  Check that $c=1+2i$ is fixed and $T(z)-c=(1+i)(z-c)$ . Also, $(1+i)^4=-4$ .
Answer

The unique solution is $z=1+(10/3)i$ .

Full solution

A fixed point satisfies $c=(1+i)c+2-i$ , so $-ic=2-i$ and $c=1+2i$ .
Subtraction gives
$T(z)-c=(1+i)(z-c).$
Repeating this relation four times yields $T^{\circ4}(z)-c=(1+i)^4(z-c)$ , where $T^{\circ4}$ means four successive applications.
Since $(1+i)^2=2i$ , its fourth power is $-4$ .
Thus
$T^{\circ4}(z)=5c-4z.$

Write $z=x+yi$ .
The equation $5(1+2i)-4(x+yi)=x-yi$ gives $5-4x=x$ and $10-4y=-y$ .
Hence $x=1$ and $y=10/3$ .
These two linear equations have unique solutions, and substitution into $5c-4z=\overline z$ verifies the result.
The fixed point is useful because the repeated affine rule becomes repeated multiplication after translation; the calculation never requires expanding the original nested expression.

Answer

The unique solution is $z=1+(10/3)i$ .

Key idea

Translating an affine complex transformation to a fixed point turns iteration into powers of one multiplier.
Problem 2: Mark complete
Problem 3 Which points make the quotient imaginary?

Difficulty: 1 of 3 stars, Stretch

For $z\ne-2$ , define $w=(z-2)/(z+2)$ . A number is called purely imaginary here when its real part is zero, including $0$ .

(a) Prove that $w$ is purely imaginary exactly when $|z|=2$ .

(b) Write every such $z$ as a formula involving one real parameter $t$ , where $w=it$ . Explain exactly which point of the circle your formula omits and why it cannot occur.
1. Hint 1
  Multiply numerator and denominator by $\overline z+2$ . The denominator is a positive real number.
2. Hint 2
  Solve $z-2=it(z+2)$ for $z$ , then multiply by the conjugate of $1-it$ .
Answer

(a) $\operatorname{Re}w=(|z|^2-4)/|z+2|^2$ . (b) $z=2(1-t^2+2it)/(1+t^2)$ for every real $t$ ; the omitted point is $-2$ .

Full solution

Multiplying by the conjugate of the denominator gives
$w=(|z|^2-4+2z-2\overline z)/|z+2|^2.$
Because $z-\overline z$ is purely imaginary and the denominator is positive for $z\ne-2$ , the real part is $(|z|^2-4)/|z+2|^2$ .
It vanishes exactly when $|z|=2$ , proving part (a).

For part (b), $z-2=it(z+2)$ gives
$z=2(1+it)/(1-it).$
The denominator never vanishes for real $t$ .
Multiplying by $1+it$ yields
$z=2(1-t^2+2it)/(1+t^2).$
The real part of this expression plus $2$ is $4/(1+t^2)>0$ , so it never equals $-2$ .
Conversely, every point on the circle other than $-2$ has a defined quotient whose real part is zero by part (a).
Taking its imaginary part as $t$ reconstructs that point by the same algebra.
The point $-2$ is absent precisely because the original quotient is undefined there.

Answer

(a) $\operatorname{Re}w=(|z|^2-4)/|z+2|^2$ . (b) $z=2(1-t^2+2it)/(1+t^2)$ for every real $t$ ; the omitted point is $-2$ .

Key idea

Conjugation can reveal the geometry of a complex quotient and produce a complete rational parametrization.
Problem 3: Mark complete
Problem 4 Two roots at one distance

Difficulty: 2 of 3 stars, Challenge

For which real numbers $k$ do the two roots of $z^2-(2+2i)z+ki=0$ have equal modulus? Count a repeated root twice. Find both roots for every permitted $k$ , and prove that no other value of $k$ works.

Builds on Complex Roots of Quadratics
1. Hint 1
  The midpoint of the roots is $1+i$ . Write the roots as $1+i+v$ and $1+i-v$ .
2. Hint 2
  Equal squared distances from the origin require $\operatorname{Re}((1+i)\overline v)=0$ . In coordinates this means that the real and imaginary parts of $v$ sum to zero.
Answer

Exactly $k\ge2$ works. With $t=\sqrt{k/2-1}$ , the roots are $(1+t)+(1-t)i$ and $(1-t)+(1+t)i$ . At $k=2$ both equal $1+i$ .

Full solution

The sum of the two roots is $2+2i$ , so write them as $u+v$ and $u-v$ , where $u=1+i$ and $v=a+bi$ for real $a,b$ .
Their squared moduli differ by
$4\operatorname{Re}(u\overline v)=4(a+b).$
Consequently they have equal modulus exactly when $b=-a$ , so $v=t(1-i)$ for a real $t$ .
This argument includes $t=0$ , corresponding to a repeated root.

The product of these roots is
$u^2-v^2=2i-(-2it^2)=2i(1+t^2).$
Comparing with the constant term gives
$k=2(1+t^2)\ge2.$
Conversely, for every $k\ge2$ , choose
$t=\sqrt{k/2-1}.$
The displayed roots have sum $2+2i$ , product $ki$ , and equal squared moduli $2+2t^2$ , so multiplying their linear factors reproduces the given quadratic.
Replacing $t$ by $-t$ merely exchanges the roots, and $t=0$ produces exactly the stated repeated-root boundary.

Answer

Exactly $k\ge2$ works. With $t=\sqrt{k/2-1}$ , the roots are $(1+t)+(1-t)i$ and $(1-t)+(1+t)i$ . At $k=2$ both equal $1+i$ .

Key idea

Equal moduli impose a perpendicularity condition on the displacement from the midpoint of two complex roots.
Problem 4: Mark complete
Problem 5 A prime real part

Difficulty: 2 of 3 stars, Challenge

Integers $a,b$ satisfy $(a+bi)^2=p+qi$ , where $p$ is a positive prime and $q$ is a positive integer.

(a) Determine every possible form of $a,b,q$ in terms of $p$ , and prove that $p$ must be odd and $q$ must be divisible by $4$ .

(b) Find all such squares with $q=60$ .
1. Hint 1
  Compare real and imaginary parts, then factor the difference of two squares in the real part.
2. Hint 2
  The two integers $a-b$ and $a+b$ have the same parity. Their positive product is prime, and $ab>0$ .
Answer

(a) $p$ is any odd prime, $q=(p^2-1)/2$ , and $(a,b)=\pm((p+1)/2,(p-1)/2)$ with the signs chosen together. (b) $p=11$ and $a+bi=\pm(6+5i)$ , whose square is $11+60i$ .

Full solution

Equating parts gives $(a-b)(a+b)=p$ and $2ab=q>0$ .
Thus $a,b$ are nonzero and have the same sign, while $a^2>b^2$ gives $|a|>|b|$ .
If both are positive, the positive factorization of the prime must be $a-b=1$ and $a+b=p$ .
Therefore $a=(p+1)/2$ and $b=(p-1)/2$ .
If both are negative, changing both signs reduces to this case without changing the square.

The factors $a-b$ and $a+b$ have the same parity, so the factorization $1\cdot2$ is impossible: $p\ne2$ .
Every odd prime gives integer $a,b$ of the displayed form and
$q=2ab=(p^2-1)/2.$
Writing $p=2m+1$ yields $q=2m(m+1)$ , which is divisible by $4$ because one of two consecutive integers is even.
This also verifies the converse construction.
Finally, $q=60$ gives $p^2=121$ , hence the positive prime $p=11$ .
The two allowed square roots are $6+5i$ and $-6-5i$ ; direct multiplication confirms their square.

Answer

(a) $p$ is any odd prime, $q=(p^2-1)/2$ , and $(a,b)=\pm((p+1)/2,(p-1)/2)$ with the signs chosen together. (b) $p=11$ and $a+bi=\pm(6+5i)$ , whose square is $11+60i$ .

Key idea

Integer factorization can control complex square roots more sharply than a general square-root formula.
Problem 5: Mark complete
Problem 6 A product of two distances

Difficulty: 2 of 3 stars, Challenge

A complex number $z$ moves on the unit circle $|z|=1$ . Determine the least and greatest possible values of $|z-1|\,|z-i|$ , and identify every point attaining each extreme. Give an exact argument without trigonometry.

Text description of this figure

The complex plane, with a horizontal real axis labeled Re, a vertical imaginary axis labeled Im, and the origin labeled 0. The unit circle is drawn centered at the origin. Three points on the circle are marked with dots: the point 1, where the circle meets the positive real axis; the point i, where it meets the positive imaginary axis; and a point z in the upper left quarter of the circle. Two straight segments join z to 1 and z to i.
1. Hint 1
  Write $z=x+yi$ . The squared distances are $2-2x$ and $2-2y$ .
2. Hint 2
  Let $s=x+y$ and use $x^2+y^2=1$ to express $xy$ in terms of $s$ . Then use $(x-y)^2\ge0$ to bound $s$ .
Answer

The minimum is $0$ , attained at $z=1$ and $z=i$ . The maximum is $2+\sqrt2$ , attained only at $z=-(1+i)/\sqrt2$ .

Full solution

Write $z=x+yi$ , so $x^2+y^2=1$ .
The square of the requested product is
$(2-2x)(2-2y)=4(1-x-y+xy).$
Set $s=x+y$ .
Since $s^2=1+2xy$ , this becomes $2(s-1)^2$ , and the nonnegative product itself is $\sqrt2\,|s-1|$ .

The identity $2-s^2=(x-y)^2\ge0$ shows
$-\sqrt2\le s\le\sqrt2.$
The largest distance from $1$ in this interval is $1+\sqrt2$ , attained only when $s=-\sqrt2$ .
Equality in the bound for $s$ then gives $x=y=-1/\sqrt2$ , so the largest product is $\sqrt2(1+\sqrt2)=2+\sqrt2$ at the claimed unique point.
The minimum cannot be below zero and is attained at $z=1$ or $z=i$ .
Conversely, a zero product requires one of its two nonnegative factors to vanish, so there are no further minimum points.
Each equality point lies on the unit circle, completing both sharpness checks.

Answer

The minimum is $0$ , attained at $z=1$ and $z=i$ . The maximum is $2+\sqrt2$ , attained only at $z=-(1+i)/\sqrt2$ .

Key idea

Products of geometric distances can simplify after squaring and replacing two coordinates by a symmetric sum.
Problem 6: Mark complete
Problem 7 A rule that stays on the circle

Difficulty: 2 of 3 stars, Challenge

For $|z|=1$ , define $T(z)=(z+2)/(2z+1)$ .

(a) Prove the denominator is never zero and $|T(z)|=1$ .

(b) Find every point on the unit circle satisfying $T(z)=z^2$ . Justify every candidate in the original quotient equation.

Builds on Complex Roots of Quadratics
1. Hint 1
  Compare $|z+2|^2$ and $|2z+1|^2$ , using $z\overline z=1$ .
2. Hint 2
  After clearing the nonzero denominator, one root is visible at $z=1$ . Verify the factorization $(z-1)(2z^2+3z+2)$ by multiplication.
Answer

The solutions in part (b) are $z=1$ and $z=(-3\pm i\sqrt7)/4$ .

Full solution

The only possible zero of $2z+1$ is $z=-1/2$ , which does not have modulus $1$ .
Thus the quotient is defined on the entire unit circle.
On that circle, $|z+2|^2=5+2(z+\overline z)$ and
$|2z+1|^2=5+2(z+\overline z).$
The denominator has positive squared modulus, so their equality proves $|T(z)|=1$ .

For part (b), clearing the verified nonzero denominator gives
$2z^3+z^2-z-2=0.$
Direct multiplication shows that the left side is $(z-1)(2z^2+3z+2)$ .
Therefore either $z=1$ or the quadratic formula gives
$z=(-3\pm i\sqrt7)/4.$
Both latter numbers have squared modulus $(9+7)/16=1$ , and $1$ also lies on the unit circle.
None makes the denominator zero.
Each solves the factored equation, so reversing the denominator-clearing step verifies the original quotient equation.
The factorization and quadratic formula exhaust all cases.

Answer

The solutions in part (b) are $z=1$ and $z=(-3\pm i\sqrt7)/4$ .

Key idea

Before solving a complex rational equation on a locus, prove its denominator and its effect on the locus are controlled.
Problem 7: Mark complete
Problem 8 Four points with two constraints

Difficulty: 3 of 3 stars, Deep challenge

Four complex numbers $z_1,z_2,z_3,z_4$ , with repetitions allowed, satisfy $|z_j|=1$ for every $j$ , $z_1+z_2+z_3+z_4=0$ , and $z_1z_2z_3z_4=1$ . Classify every possible collection of four numbers, disregarding their order. Prove your description is necessary and sufficient, including repeated-point cases.
1. Hint 1
  First prove that four points on the unit circle with sum zero can be grouped into opposite pairs. Compare the pairs $z_1,z_2$ and $-z_3,-z_4$ , which have the same midpoint.
2. Hint 2
  A chord of a circle is determined by its nonzero midpoint: its endpoints lie on the line through that midpoint perpendicular to the radius. If the midpoint is zero, the endpoints are already opposites.
Answer

Exactly the collections $\{u,-u,\overline u,-\overline u\}$ occur, where $|u|=1$ ; repetitions are retained.

Full solution

We first establish the opposite-pair claim in the hint.
The two pairs $z_1,z_2$ and $-z_3,-z_4$ have the same midpoint $m$ .
If $m=0$ , both pairs already consist of opposites.
If $m\ne0$ , write either pair as $m+v,m-v$ .
Equal unit moduli give $\operatorname{Re}(m\overline v)=0$ and
$|m|^2+|v|^2=1.$
The first condition restricts $v$ to the line through the origin perpendicular to $m$ , and the second fixes its length.
Thus there are only the two choices $v$ and $-v$ , or only $v=0$ if $|m|=1$ .
The two unordered pairs must coincide.
In either case the original collection can be written $\{u,-u,v,-v\}$ with $|u|=|v|=1$ .

The product condition is now $u^2v^2=1$ , so $(uv-1)(uv+1)=0$ and $uv=1$ or $uv=-1$ .
Since $1/u=\overline u$ for a unit complex number, $v=\overline u$ or
$v=-\overline u.$
Both choices give the same collection $\{u,-u,\overline u,-\overline u\}$ .
Conversely, every such collection has unit moduli, zero sum, and product
$u^2\overline u^{\,2}=1.$
The proof allowed zero chord length and coinciding points, so no repeated-point exceptions are lost.

A nonzero midpoint fixes the unordered chord.

Answer

Exactly the collections $\{u,-u,\overline u,-\overline u\}$ occur, where $|u|=1$ ; repetitions are retained.

Key idea

A zero-sum condition on a circle can force opposite pairing; an additional product then restricts the remaining geometry.
Problem 8: Mark complete
Problem 9 A nonlinear distance condition

Difficulty: 3 of 3 stars, Deep challenge

A complex number $z$ satisfies $|z|=|z-1|\,|z+1|$ . Find the greatest possible imaginary part of $z$ , and determine all numbers attaining it. Also prove that no number on the imaginary axis satisfies the condition.

Builds on Completing the Square
1. Hint 1
  Write $z=x+yi$ and set $t=|z|^2=x^2+y^2$ . Square the distance condition and use $|z-1|\,|z+1|=|z^2-1|$ .
2. Hint 2
  You should obtain $4y^2=-t^2+3t-1$ and $4x^2=t^2+t+1$ . Complete the square in the expression for $y^2$ .
Answer

The greatest imaginary part is $\sqrt5/4$ , attained exactly at $z=(\sqrt{19}+i\sqrt5)/4$ and $z=(-\sqrt{19}+i\sqrt5)/4$ . There are no solutions with real part zero.

Full solution

Write $z=x+yi$ and
$t=x^2+y^2\ge0.$
Squaring the nonnegative modulus equation gives
$t=|z^2-1|^2=t^2-2x^2+2y^2+1.$
Substituting $x^2=t-y^2$ yields
$4y^2=-t^2+3t-1=5/4-(t-3/2)^2.$
Therefore $y^2\le5/16$ , which implies
$y\le\sqrt5/4.$
Equality in this upper bound requires $t=3/2$ and
$y=\sqrt5/4.$
It then forces $x^2=t-y^2=19/16$ , giving precisely the two displayed numbers.

For either candidate, $x^2+y^2=3/2$ and the displayed squared equation holds.
Its two original sides are nonnegative, so equality of their squares implies the required equality of moduli; no sign condition was lost.
Finally, the same relation gives
$4x^2=4t-4y^2=t^2+t+1.$
This is strictly positive for every $t\ge0$ , ruling out $x=0$ .
Introducing the squared modulus reduced the apparently quartic condition to a quadratic bound with explicitly checked equality points.

Answer

The greatest imaginary part is $\sqrt5/4$ , attained exactly at $z=(\sqrt{19}+i\sqrt5)/4$ and $z=(-\sqrt{19}+i\sqrt5)/4$ . There are no solutions with real part zero.

Key idea

For nonlinear distance constraints, the squared modulus can be a useful new variable that exposes a sharp quadratic bound.
Problem 9: Mark complete
Problem 10 Can a rational point return to one?

Difficulty: 3 of 3 stars, Deep challenge

Let $z$ have rational real and imaginary parts and satisfy $|z|=1$ . Classify all such $z$ for which $z^n=1$ for at least one positive integer $n$ . In particular, decide whether any positive power of $(3+4i)/5$ equals $1$ . Prove the classification without trigonometric complex form or a theorem about roots of unity.
1. Hint 1
  Consider the real quantities $S_j=z^j+\overline z^{\,j}$ . With $t=z+\overline z$ , multiplication gives $S_{j+1}=tS_j-S_{j-1}$ .
2. Hint 2
  Write $t=A/B$ in lowest terms, with $B>0$ , and set $N_j=B^jS_j$ . Prove that $N_j$ is an integer and that $N_j-A^j$ is divisible by $B$ for every $j\ge1$ . What happens if $S_n=2$ ?
Answer

Exactly $z=1,-1,i,-i$ work. No positive power of $(3+4i)/5$ equals $1$ .

Full solution

Because $|z|=1$ , we have
$z\overline z=1.$
Set $t=z+\overline z=2\operatorname{Re}z$ , a rational number, and
$S_j=z^j+\overline z^{\,j}.$
Multiplication gives $S_0=2$ , $S_1=t$ , and
$S_{j+1}=tS_j-S_{j-1}.$
Write $t=A/B$ in lowest terms with integers $A,B$ and $B>0$ .
The scaled values $N_j=B^jS_j$ satisfy $N_0=2$ , $N_1=A$ , and
$N_{j+1}=AN_j-B^2N_{j-1}.$
Thus they are integers.
Induction from $N_1=A$ shows that $N_j-A^j$ is divisible by $B$ for every $j\ge1$ .

If $z^n=1$ , then also $\overline z^{\,n}=1$ , so $N_n=2B^n$ is divisible by $B$ .
Hence $B$ divides $A^n$ .
Since $A$ and $B$ are coprime, their positive powers remain coprime to $B$ , forcing $B=1$ .
Therefore $t$ is an integer.
Also $|\operatorname{Re}z|\le1$ , so $t$ is one of $-2,-1,0,1,2$ .
The choices $t=\pm1$ would give imaginary part $\pm\sqrt3/2$ , which is irrational: if $\sqrt3=a/b$ in lowest terms, $a^2=3b^2$ forces both $a$ and $b$ divisible by $3$ , a contradiction.
The remaining choices give exactly $z=\pm1,\pm i$ , whose powers do reach $1$ .
The point $(3+4i)/5$ is absent from this list.

Answer

Exactly $z=1,-1,i,-i$ work. No positive power of $(3+4i)/5$ equals $1$ .

Key idea

A recurrence for conjugate power sums can turn a complex periodicity question into a denominator obstruction.
Problem 10: Mark complete