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Complex Numbers and the Quadratic Formula: 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.

0 of 10 completed

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Problem 1 of 10

Problem 1 Large powers without expansion

Difficulty: 1 of 3 stars, Stretch

Let $i^2=-1$ . For a positive integer $n$ , define
$Z_n=\frac{(1+i)^{n+2}}{(1-i)^n}.$
Find a simple expression for $Z_n$ . Determine exactly which positive integers $n$ make $Z_n$ a positive real number, and give its value in those cases.
1. Hint 1
  Separate two factors of $1+i$ , and simplify the quotient $(1+i)/(1-i)$ .
2. Hint 2
  The powers of $i$ repeat every four exponents. Keep the two extra factors when deciding the exponent.
Answer

$Z_n=2i^{n+1}$ . It is positive real exactly when $n=4k+3$ for an integer $k\ge0$ , and then $Z_n=2$ .

Full solution

The denominator is nonzero, so the quotient can be reorganized as
$Z_n=(1+i)^2\left(\frac{1+i}{1-i}\right)^n.$
First,
$(1+i)^2=1+2i+i^2=2i.$
Rationalizing the other quotient gives
$(1+i)/(1-i)=(1+i)^2/((1-i)(1+i))=2i/2=i.$
Therefore
$Z_n=2i^{n+1}.$

The powers of $i$ cycle through $1,i,-1,-i$ , repeating every four exponents because $i^4=1$ .
Thus the only possible values of $Z_n$ are $2,2i,-2,-2i$ .
Its value is positive real precisely when $i^{n+1}=1$ , which requires $n+1$ to be divisible by $4$ .

Among positive integers this means $n=3,7,11,\ldots$ , equivalently $n=4k+3$ with $k\ge0$ .
The value is $2$ in every such case.
The quotient structure eliminates the large powers before any lengthy multiplication is needed.

Answer

$Z_n=2i^{n+1}$ . It is positive real exactly when $n=4k+3$ for an integer $k\ge0$ , and then $Z_n=2$ .

Key idea

Simplify a common base ratio before expanding powers, then use the short cycle of powers of $i$ .
Problem 1: Mark complete
Problem 2 How wide is the framed sign?

Difficulty: 1 of 3 stars, Stretch

A rectangular sign has width $w$ meters and length $w+6$ meters, where $w>0$ . A frame of uniform width $2$ meters surrounds the sign on all four sides, with square outer corners. The frame area alone is half the sign area.

Find the exact dimensions of the sign. Explain why the other quadratic root cannot represent a sign, and verify the area relationship.

Builds on Completing the Square, Applications of Quadratics
1. Hint 1
  The outer dimensions increase by $4$ meters in each direction, not by $2$ . Subtract the sign area to obtain the frame area.
2. Hint 2
  The area condition leads to $w^2-10w-80=0$ . Complete the square or use the quadratic formula, then impose $w>0$ .
Answer

Width $5+\sqrt{105}$ meters; length $11+\sqrt{105}$ meters.

Full solution

The sign area is $w(w+6)$ and the outer rectangle has dimensions $w+4$ and $w+10$ .
Thus the frame area is
$(w+4)(w+10)-w(w+6)=8w+40.$
The stated ratio gives $8w+40=\frac12w(w+6)$ , or
$w^2-10w-80=0.$

Completing the square yields $(w-5)^2=105$ , so
$w=5\pm\sqrt{105}.$
Since $\sqrt{105}>5$ , the minus choice is negative and violates the required positive width.
The valid width is $5+\sqrt{105}$ and the length is $11+\sqrt{105}$ .
Both outer dimensions are then positive as well.

For a check without a long radical expansion, the quadratic equation gives
$w^2=10w+80.$
Hence the sign area is $w^2+6w=16w+80$ , while the frame area is $8w+40$ , exactly half as large.
This verifies the original geometric condition using the equation satisfied by the chosen positive root.

Answer

Width $5+\sqrt{105}$ meters; length $11+\sqrt{105}$ meters.

Key idea

Model the physical regions before solving; a quadratic equation may retain roots that the geometry excludes.
Problem 2: Mark complete
Problem 3 Extracting a complex square root

Difficulty: 1 of 3 stars, Stretch

Find all complex numbers $z=a+bi$ , with $a,b$ real, satisfying $z^2=7+24i$ . Do not assume either $a$ or $b$ is positive. Prove that all sign possibilities have been handled.

Builds on Squares of Binomials
1. Hint 1
  Equate real and imaginary parts after squaring $a+bi$ .
2. Hint 2
  Use $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2$ to find $a^2+b^2$ . Then recover the two squares and their compatible signs.
Answer

$z=4+3i$ or $z=-4-3i$ .

Full solution

Expanding gives
$z^2=(a^2-b^2)+2ab\,i.$
Therefore the required conditions are $a^2-b^2=7$ and $2ab=24$ , so $ab=12$ .

Rather than substituting $b=12/a$ into a fourth-degree equation, combine the two known quantities:
$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=7^2+24^2=625.$
Since $a^2+b^2\ge0$ , this implies
$a^2+b^2=25.$
Adding and subtracting the equations for the sum and difference of squares gives $a^2=16$ and $b^2=9$ .

Thus $a=\pm4$ and $b=\pm3$ .
The condition $ab=12>0$ requires their signs to match, leaving $(4,3)$ and $(-4,-3)$ .
These give $4+3i$ and its negative.
Direct squaring yields $7+24i$ in both cases.
The mixed-sign possibilities instead give imaginary part $-24$ , so they do not solve the original equation.

Answer

$z=4+3i$ or $z=-4-3i$ .

Key idea

For a complex square root, recover a sum of squares before choosing signs for the real and imaginary parts.
Problem 3: Mark complete
Problem 4 When a quadratic becomes linear

Difficulty: 2 of 3 stars, Challenge

For real $t$ , consider
$(t-1)x^2-2tx+t+1=0.$
Find every $t$ for which this equation has exactly one positive real solution, and give that solution. Classify the number of positive real solutions for all other $t$ as well. Count distinct solutions, and handle parameters for which the equation is not quadratic.

Builds on The Quadratic Formula, Factoring Quadratics, Inequality Basics
1. Hint 1
  Check the leading coefficient before dividing by it. Also test whether $x=1$ is always a root.
2. Hint 2
  Factor the expression as $(x-1)((t-1)x-(t+1))$ . For $t\ne1$ , determine the sign of $(t+1)/(t-1)$ .
Answer

Exactly one positive solution, $x=1$ , when $-1\le t\le1$ . For $t<-1$ or $t>1$ , there are two distinct positive solutions: $1$ and $(t+1)/(t-1)$ .

Full solution

The expression factors as $(x-1)((t-1)x-(t+1))$ , so $x=1$ is always a solution.
At $t=1$ , the original equation is linear: $-2x+2=0$ .
It has exactly the positive solution $1$ ; a quadratic formula with denominator $2(t-1)$ would not apply here.

For $t\ne1$ , the other root is
$r=(t+1)/(t-1).$
It never equals $1$ , since that would require $t+1=t-1$ .
Thus there is a second positive solution exactly when this quotient is positive.

If $t>1$ , its numerator and denominator are positive.
If $t<-1$ , both are negative.
These two ranges give a second positive root.
At $t=-1$ , the second root is $0$ , which is not positive.
For $-1<t<1$ , the numerator is positive and the denominator negative, so the second root is negative.
Including the linear case $t=1$ , exactly one positive solution occurs for the whole interval $[-1,1]$ .
There are never zero positive solutions because $1$ always works.

Answer

Exactly one positive solution, $x=1$ , when $-1\le t\le1$ . For $t<-1$ or $t>1$ , there are two distinct positive solutions: $1$ and $(t+1)/(t-1)$ .

Key idea

A parameter can lower the degree; separate that case before applying a quadratic formula or dividing by a coefficient.
Problem 4: Mark complete
Problem 5 Recover a flight from two measurements

Difficulty: 2 of 3 stars, Challenge

A ball follows the height model $h=-5t^2+vt+h_0$ , where $h$ is in meters, $t$ is time in seconds after release, and $v,h_0$ are unknown real constants. The model applies for $t\ge0$ until the ball first reaches the ground. The ball is $30$ meters high at $t=1$ and $50$ meters high at $t=3$ .

(a) Determine $v$ and $h_0$ .

(b) Prove the ball never exceeds $50$ meters during this flight, and find when it reaches that height.

(c) Find the exact time when it first reaches the ground.

Builds on Completing the Square, Applications of Quadratics
1. Hint 1
  Use the two measurements to form a linear system for $v,h_0$ .
2. Hint 2
  Once the model is known, write it as a constant minus a square. The same form settles both the greatest height and the ground time.
Answer

(a) $v=30$ , $h_0=5$ . (b) Greatest height $50$ meters at $t=3$ seconds. (c) Ground time $3+\sqrt{10}$ seconds.

Full solution

At $t=1$ , the measurement gives $-5+v+h_0=30$ , hence $v+h_0=35$ .
At $t=3$ , it gives $-45+3v+h_0=50$ , hence $3v+h_0=95$ .
Subtracting yields $2v=60$ , so $v=30$ and $h_0=5$ .

The model is therefore
$h=-5t^2+30t+5=50-5(t-3)^2.$
A square is nonnegative, so $h\le50$ for every real $t$ and therefore throughout the flight.
Equality occurs exactly when $t=3$ .
This time lies in the physical flight because the given measurement places the ball above ground then.

For the ground time, set $h=0$ .
The square form gives $(t-3)^2=10$ , hence
$t=3\pm\sqrt{10}.$
Since $\sqrt{10}>3$ , the minus root is negative and lies before release.
The plus root is positive.
There is no other nonnegative root, so the ball first reaches the ground at $3+\sqrt{10}$ seconds.
The height remains positive between release and that root, as the square form also confirms.

Answer

(a) $v=30$ , $h_0=5$ . (b) Greatest height $50$ meters at $t=3$ seconds. (c) Ground time $3+\sqrt{10}$ seconds.

Key idea

Recover unknown model parameters with linear equations, then use a square form to interpret the quadratic.
Problem 5: Mark complete
Problem 6 A nonreal number with an integer sum

Difficulty: 2 of 3 stars, Challenge

Find all nonreal complex numbers $z$ for which $z+1/z$ is an integer. Here an integer means a real integer. Prove that your list is complete.

Builds on The Quadratic Formula
1. Hint 1
  Write $z=a+bi$ with $b\ne0$ , and express $1/z$ using the conjugate.
2. Hint 2
  The imaginary part can vanish only if $a^2+b^2=1$ . The real value is then $2a$ ; nonzero $b$ puts strict bounds on it.
Answer

$z=\pm i$ , $z=(1\pm i\sqrt3)/2$ , or $z=(-1\pm i\sqrt3)/2$ .

Full solution

Write $z=a+bi$ with real $a,b$ and $b\ne0$ .
In particular, $z\ne0$ , so the reciprocal is defined.
Rationalizing gives
$1/z=(a-bi)/(a^2+b^2).$
Thus the imaginary part of $z+1/z$ is $b(1-1/(a^2+b^2))$ .
Because $b\ne0$ , a real sum requires $a^2+b^2=1$ .

Under that condition, the real part of the sum is $2a$ .
Also, $b^2>0$ forces $a^2<1$ , so $-2<2a<2$ .
The only integers in this open interval are $-1,0,1$ .

If the integer is $0$ , then $a=0$ and $b=\pm1$ , giving $\pm i$ .
If it is $1$ , then $a=1/2$ and
$b=\pm\sqrt3/2.$
If it is $-1$ , then $a=-1/2$ and
$b=\pm\sqrt3/2.$
Each listed number is nonreal and has $a^2+b^2=1$ , so substitution into the reciprocal formula confirms its stated integer sum.
The strict bounds exclude the real boundary values $z=1,-1$ .

Answer

$z=\pm i$ , $z=(1\pm i\sqrt3)/2$ , or $z=(-1\pm i\sqrt3)/2$ .

Key idea

Requiring a complex expression to be real can impose a strong equation before any integer cases are considered.
Problem 6: Mark complete
Problem 7 A quadratic inside a reciprocal equation

Difficulty: 2 of 3 stars, Challenge

Find all real solutions of
$x^2+\frac1{x^2}-3\left(x+\frac1x\right)-2=0.$
Explain why a solution of an intermediate quadratic need not correspond to a real value of $x$ .

Builds on The Quadratic Formula, Squares of Binomials, Algebraic Fractions
1. Hint 1
  The domain excludes $x=0$ . Let $u=x+1/x$ , and express $x^2+1/x^2$ in terms of $u$ .
2. Hint 2
  The equation in $u$ factors. For each resulting value, solve $x^2-ux+1=0$ and inspect its discriminant.
Answer

$x=2+\sqrt3$ or $x=2-\sqrt3$ .

Full solution

The original equation requires $x\ne0$ .
Put $u=x+1/x$ .
Squaring gives $x^2+1/x^2=u^2-2$ , so the equation becomes
$u^2-3u-4=0.$
Factoring gives $(u-4)(u+1)=0$ , hence $u=4$ or $u=-1$ .

For $u=4$ , multiplying $x+1/x=4$ by the nonzero $x$ yields
$x^2-4x+1=0.$
The quadratic formula gives
$x=(4\pm\sqrt{16-4})/2=2\pm\sqrt3.$
Both are nonzero, and their defining quadratic shows $x+1/x=4$ , so both satisfy the original equation.

For $u=-1$ , the same multiplication yields $x^2+x+1=0$ .
Its discriminant is $1-4=-3<0$ , so it has no real solutions.
Thus this intermediate value cannot come from a real $x$ .
Substitution into a new variable simplifies the equation, but back-substitution is necessary to test which values of that variable are actually attainable in the original domain.

Answer

$x=2+\sqrt3$ or $x=2-\sqrt3$ .

Key idea

After a substitution, solve back in the original variable; admissible intermediate values are not automatic.
Problem 7: Mark complete
Problem 8 Real roots but no rational roots

Difficulty: 3 of 3 stars, Deep challenge

For a real parameter $t$ , consider
$x^2-2(t+1)x+t^2-3t+4=0.$
(a) Find exactly when the equation has two distinct positive real roots. Also describe the boundary case where there is one repeated real root.

(b) Prove that whenever $t$ is an integer and the roots are real, both roots are irrational.

Builds on Completing the Square, The Quadratic Formula, Sums and Products of Roots
1. Hint 1
  Complete the square in $x$ . To check positivity of real roots, use their sum and product rather than comparing two radicals directly.
2. Hint 2
  For integer $t$ , rationality would make $\sqrt{5t-3}$ rational. A rational square root of an integer must be an integer; compare integer squares with their remainders on division by $5$ .
Answer

(a) Two distinct positive roots exactly when $t>3/5$ ; at $t=3/5$ the repeated root is $8/5$ ; for $t<3/5$ there are no real roots. (b) For every integer $t\ge1$ , both roots are irrational.

Full solution

Completing the square gives
$(x-(t+1))^2=5t-3.$
Thus real roots exist exactly when $t\ge3/5$ .
Equality gives the repeated root $t+1=8/5$ ; a strict inequality gives two distinct roots $t+1\pm\sqrt{5t-3}$ .

For $t\ge3/5$ , the root sum $2(t+1)$ is positive.
Their product is
$t^2-3t+4=(t-3/2)^2+7/4>0.$
Real roots with positive product have the same nonzero sign, and their positive sum forces both to be positive.
This proves part (a).

For part (b), real roots at integer $t$ require $t\ge1$ .
If either root were rational, then $\sqrt{5t-3}$ would be rational.
A rational number whose square is an integer is itself an integer: writing it as a reduced fraction $p/q$ , any prime divisor of $q$ would divide $p^2$ and hence $p$ , contradicting reduction.
Thus we would have an integer $m$ with $m^2=5t-3$ .

But the square of an integer has remainder $0,1,$ or $4$ on division by $5$ , as squaring the remainders $0,1,2,3,4$ shows.
The number $5t-3$ has remainder $2$ .
This contradiction rules out rationality of either root.

Answer

(a) Two distinct positive roots exactly when $t>3/5$ ; at $t=3/5$ the repeated root is $8/5$ ; for $t<3/5$ there are no real roots. (b) For every integer $t\ge1$ , both roots are irrational.

Key idea

Separate real existence, root signs, and arithmetic type; each calls for a different short argument.
Problem 8: Mark complete
Problem 9 Two quadratics that never go negative

Difficulty: 3 of 3 stars, Deep challenge

Find all ordered pairs of integers $(a,b)$ such that both
$x^2+ax+b\quad\text{and}\quad x^2+bx+a$
are nonnegative for every real number $x$ . Prove that your finite list is complete.

Builds on Completing the Square, Inequality Basics
1. Hint 1
  Complete the square in each expression to turn the condition for every $x$ into two inequalities in $a,b$ .
2. Hint 2
  Both integers are nonnegative. If $a$ is the larger one, use $a^2\le4b\le4a$ to bound it; then make a short complete case table.
Answer

$(a,b)=(0,0),(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)$ .

Full solution

Completing the square gives
$x^2+ax+b=(x+a/2)^2+b-a^2/4.$
This is nonnegative for every real $x$ exactly when $a^2\le4b$ : sufficiency follows from the square, and necessity follows by choosing $x=-a/2$ .
Similarly, the second expression requires $b^2\le4a$ .

These inequalities force $a,b\ge0$ .
If the larger of the two is $a>0$ , then $a^2\le4b\le4a$ , so $a\le4$ .
The same argument applies with $a,b$ reversed.
If the larger is zero, both are zero.
Thus in every case $a,b\in\{0,1,2,3,4\}$ .

Now impose both inequalities for each possible $a$ .
For $a=0$ , only $b=0$ works.
For $a=1$ , they require
$1\le b\le2.$
For $a=2$ , again $b=1$ or $2$ .
For $a=3$ , $b\ge9/4$ and $b^2\le12$ force $b=3$ .
For $a=4$ , $b\ge4$ and $b\le4$ force $b=4$ .

These seven pairs satisfy both inequalities, so their completed-square forms verify nonnegativity for every real $x$ .
The bound and the complete case list exclude all other integer pairs.

Answer

$(a,b)=(0,0),(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)$ .

Key idea

A universal quadratic condition can become a finite integer problem once completing the square supplies bounds.
Problem 9: Mark complete
Problem 10 When a complex quotient keeps integer parts

Difficulty: 3 of 3 stars, Deep challenge

Find all complex numbers $z=a+bi$ , where $a,b$ are integers and $z\ne1$ , for which
$(z+1)/(z-1)$
also has integer real and imaginary parts. Prove that your list is complete; $z$ is allowed to be real.

Builds on Difference of Squares
1. Hint 1
  Write the quotient as $1+2/(z-1)$ , and put $z-1=u+vi$ with integers $u,v$ .
2. Hint 2
  If $2/(u+vi)=c+di$ has integer parts, multiply squared real-and-imaginary lengths to obtain $(u^2+v^2)(c^2+d^2)=4$ . This gives a small bound for $u,v$ .
Answer

$z\in\{-1,0,2,3,\ i,-i,\ 1+i,1-i,\ 1+2i,1-2i,\ 2+i,2-i\}$ .

Full solution

Since $(z+1)/(z-1)=1+2/(z-1)$ , its parts are integers exactly when those of $2/(z-1)$ are integers.
Put $z-1=u+vi$ , where $u,v$ are integers and not both zero.
Suppose $2/(u+vi)=c+di$ with integer $c,d$ .
This quotient is nonzero, so
$c^2+d^2\ge1.$

Multiplying $(u+vi)(c+di)=2$ and adding the squares of its real and imaginary parts yields
$(u^2+v^2)(c^2+d^2)=4.$
Therefore
$1\le u^2+v^2\le4.$
The integer pairs in this range are $(\pm1,0),(0,\pm1),(\pm1,\pm1),(\pm2,0),(0,\pm2)$ .
There are no pairs with square sum $3$ , since squares at most $3$ are only $0$ and $1$ .

Every listed pair actually works.
Indeed,
$\frac2{u+vi}=\frac{2u}{u^2+v^2}-\frac{2v}{u^2+v^2}i.$
For square sum $1$ the two parts are integers; for sum $2$ they are $u,-v$ ; and for sum $4$ , one of $u,v$ is $\pm2$ and the other zero, again giving integers.

Finally, translate back using $z=1+u+vi$ .
The twelve resulting numbers are exactly those in the answer.
None equals $1$ , and the bound proves that no other integer-part complex number is possible.

Answer

$z\in\{-1,0,2,3,\ i,-i,\ 1+i,1-i,\ 1+2i,1-2i,\ 2+i,2-i\}$ .

Key idea

Multiplying sums of two squares can bound an infinite search for complex numbers with integer parts.
Problem 10: Mark complete