Complex Numbers and Equations

In the Chinese book: 复数与方程

|z−a| is a distance: modulus equations are curves; non-real roots pair up.

📖 Chapter 3 of 《复数与向量》M1 Reduction line

Try this

Choose “circle”: $∣ z - z_{1} ∣ = k$ . $∣ z - z_{1} ∣$ measures the distance between which two points?
Choose “bisector”: $P$ is equally far from $Z_{1}$ and $Z_{2}$ . What do the stamped points line up into?
Choose “ellipse” with $k = 3$ : the distances from $P$ to the two points add up to $6$ . Then set $k$ to $2$ and to $1.5$ , and click “Show the locus” to see what happens.
Choose “hyperbola”: the absolute value of the difference of the distances is $2 k$ . Why does the locus vanish when $k$ is too large?

There is only one key sentence: $∣ z_{1} - z_{2} ∣$ is the distance between the points $Z_{1}$ and $Z_{2}$ of the complex plane. So an equation about moduli is a condition about distances, which is a curve:

Equation	Locus
$∣ z - z_{0} ∣ = r$ ( $r > 0$ )	the circle with center $Z_{0}$ and radius $r$
$∣ z - z_{1} ∣ = ∣ z - z_{2} ∣$	the perpendicular bisector of the segment $Z_{1} Z_{2}$
$∣ z - z_{1} ∣ + ∣ z - z_{2} ∣ = 2 a$ , $2 a > ∣ Z_{1} Z_{2} ∣$	the ellipse with foci $Z_{1}, Z_{2}$
$∣ z - z_{1} ∣ - ∣ z - z_{2} ∣ = 2 a$ , $0 < 2 a < ∣ Z_{1} Z_{2} ∣$	the hyperbola with foci $Z_{1}, Z_{2}$

Replace the equals sign by an inequality and you get a region: $∣ z - z_{0} ∣ < r$ is the inside of the circle, and $Re z > 0$ is the right half-plane.

Predict first

The locus of $∣ z + 2∣ + ∣ z - 2∣ = 6$ is an ellipse. How far from the origin is its point nearest to the origin? (Give two decimal places.)

Choose an answer first; then you can see the result.

Why non-real roots come in pairs

Now turn it around and let complex numbers be the solutions of an equation. The quadratic $z^{2} + p z + q = 0$ with real coefficients has no real root when $Δ = p^{2} - 4 q < 0$ , but it has two roots among the complex numbers:

z = \frac{- p \pm 4 q - p ^{2} i}{2} .

Non-real roots of a real equation come in conjugate pairs

Step 1 / 4

An example
$z^{2} - 2 z + 5 = 0$ : $Δ = 4 - 20 = - 16 < 0$ . The two roots are $1 \pm 2 i$ , symmetric about the real axis.

$z^{2} - 2 z + 5 = 0 \Rightarrow z = 1 + 2 i, 1 - 2 i$
Take the conjugate of the whole equation
So z̄₀ is a root too
Both roots lie on one circle

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: the two non-real roots are conjugates of each other

As long as $Δ < 0$ , the two roots are symmetric about the real axis, and “Conjugate pair?” stays ✓.

Practice

Exercise 1Basic

The complex number $z$ satisfies $∣ z - 3 i ∣ = 2$ . Find the maximum of $∣ z ∣$ .

Exercise 2Basic

The equation $x^{2} + b x + c = 0$ with real coefficients has $1 - 2 i$ as a root. Find $b + c$ .

Exercise 3Advanced

The points $z$ with $∣ z - 4∣ = ∣ z - 2 i ∣$ form a line. Find the $x$ -coordinate of the point where it meets the real axis.

Exercise 4Challenge

The complex number $z$ satisfies $∣ z + 3∣ + ∣ z - 3∣ = 10$ . Find the minimum of $∣ z ∣$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Is a constant sum always an ellipse?

∣ z - 1∣ + ∣ z + 1∣ = 2

is the ellipse with foci

\pm 1

What went wrong? Think first, then open

Diagnosis The distance between the two fixed points is $2$ , and the sum of the distances is also $2$ . Two sides of a triangle add up to more than the third side, with equality only when $P$ lies on the segment, so the locus is just a segment.

Compare

2 k

with

∣ Z_{1} Z_{2} ∣

: if it is larger, an ellipse; if equal, the segment

Z_{1} Z_{2}

(here the part

- 1 \leq x \leq 1

of the real axis); if smaller, the empty set.

“Compare the constant with the distance between the two fixed points before you conclude anything”

Exhibit

Using the discriminant to test for real roots when the coefficients are not real

The equation

x^{2} - (2 + i) x + 2 i = 0

has discriminant

Δ = 3 - 4 i

, which is not a non-negative real number, so it has no real root.

What went wrong? Think first, then open

Diagnosis “If $Δ \geq 0$ there are real roots” was derived for equations with real coefficients. When the coefficients contain $i$ , the discriminant cannot even be “greater than zero”. In fact, substitute $x = 2$ : $4 - 4 - 2 i + 2 i = 0$ , so there is a real root.

Substitute a real root

x

and separate real and imaginary parts:

(x^{2} - 2 x) + (2 - x) i = 0

, so

x^{2} - 2 x = 0

and

x = 2

. The real root is

2

(the other root is

i

“The discriminant is only for real coefficients; with complex coefficients, turn the complex into the real first”