Modulus and Argument (II)

In one sentence

The modulus obeys the triangle inequality, so it can be bounded above and below:

∣ z_{1} ∣ - ∣ z_{2} ∣ \leq ∣ z_{1} \pm z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣,

The argument tells us the direction: if a group of complex numbers all lie in one half-plane bounded by a line through the origin, their sum cannot be $0$ . Almost every problem in this chapter uses these two moves in different places.

First, the most typical use: the roots of a polynomial cannot be too large. Let $p (z) = z^{n} + a_{n - 1} z^{n - 1} + \dots + a_{0}$ and $A = max ∣ a_{i} ∣$ .

Every root satisfies |z| < 1 + A

Step 1 / 4

Suppose there is a large root
$p (z) = z^{3} - z^{2} + 2$ , $A = 2$ , $R = 3$ . Argue by contradiction: suppose some root $z_{0}$ has $∣ z_{0} ∣ \geq 3$ . In the figure, put the test point outside the circle.

$z_{0}^{n} = - (a_{n - 1} z_{0}^{n - 1} + \dots + a_{0})$
Take moduli and enlarge the right side
Enlarge once more
Contradiction

Predict first

$p (z) = z^{4} + 3 z^{2} - 2$ . By the result above, all its roots have modulus less than what number?

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

Stop 1 · Extreme values: the modulus is a distance

“Given $∣ z ∣ = 1$ , find the maximum of some modulus” is the most common type of problem: either use the triangle inequality, or read the modulus as a distance.

One problem, several solutions

The complex number $z$ satisfies $∣ z ∣ = 1$ . Find the maximum of $∣ z - 3 - 4 i ∣$ .

Triangle inequality

$∣ z - (3 + 4 i) ∣ \leq ∣ z ∣ + ∣3 + 4 i ∣ = 1 + 5 = 6$ . Equality needs $z$ to point in the same direction as $- (3 + 4 i)$ , that is, $z = - \frac{3}{5} - \frac{4}{5} i$ , and this has $∣ z ∣ = 1$ as required. The maximum is $6$ .

CostOne line, but you must check separately that equality can be reached.

Read it as a distance

$z$ is on the unit circle, and $∣ z - (3 + 4 i) ∣$ is its distance from the point $(3, 4)$ . The center is at distance $5$ from $(3, 4)$ , so the distance lies between $5 - 1 = 4$ and $5 + 1 = 6$ ; the maximum is $6$ .

(The point

(3, 4)

does not fit in the picture, so only the unit circle is drawn.) Drag

P

to around

(- 0.6, - 0.8)

in the third quadrant: that is the point farthest from

(3, 4)

CostYou need a picture, but you see the maximum and the minimum at once.

Stop 2 · Roots of equations: the leading term beats everything

This is the proof above. Its skeleton: put one term alone on one side of the equation, take moduli, enlarge the other side with the triangle inequality, and reach a contradiction. The same move proves that if every $∣ a_{i} ∣ < 1$ , then all roots of $z^{n} + a_{n - 1} z^{n - 1} + \dots + a_{0} = 0$ satisfy $∣ z ∣ < 2$ (just take $A < 1$ ).

Stop 3 · Arguments: herding into a half-plane

If the arguments of $z_{1}, \dots, z_{n}$ all lie in some open interval $(α, α + π)$ , then they are all on the same side of a line through the origin, and so is their sum. Therefore

z_{1} + z_{2} + \dots + z_{n} \neq = 0.

Used the other way round: if $z_{1} + \dots + z_{n} = 0$ , these complex numbers cannot all be squeezed into one half-plane: every line through the origin has some of them on each side. Below, drag the arrows and try to make all three point into the right half-plane while their sum is zero. You cannot.

When the chain closes up, some part of it always has to “walk back”.

Practice

Exercise 1Basic

The complex number $z$ satisfies $∣ z ∣ = 1$ . Find the maximum of $∣ z + 3 - 4 i ∣$ .

Exercise 2Basic

The complex number $z$ satisfies $∣ z - 2 i ∣ = 1$ . Find the minimum of $∣ z ∣$ .

Exercise 3Advanced

By the method of this page, all roots of $z^{5} + 3 z^{2} - 2 = 0$ have modulus less than which number $R$ ? (Enter the value of $1 + max ∣ a_{i} ∣$ .)

Exercise 4Challenge

$∣ z_{1} ∣ = ∣ z_{2} ∣ = 1$ and $∣ z_{1} + z_{2} ∣ = 3$ . Find $∣ z_{1} - z_{2} ∣$ .

In one sentence

Every root satisfies |z| < 1 + A

Stop 1 · Extreme values: the modulus is a distance

Triangle inequality

Read it as a distance

Stop 2 · Roots of equations: the leading term beats everything

Stop 3 · Arguments: herding into a half-plane

Practice

Mistake Museum · Exhibits from this stop

Treating the modulus of a sum as the sum of the moduli

Remembering the equality condition as “equal”