Further complex numbers 1 AI HL

IB Mathematics: Applications and Interpretation · AI HL

Lesson 1.13 · Further Complex Numbers

Use polar and exponential forms to describe magnitude, angle, multiplication and phase.

Faith connection · God’s love

A complex number can be represented in several forms, each revealing something useful. Scripture describes the breadth of God’s love and points us to Jesus Christ: “that ye … may be able to comprehend … what is the breadth, and length, and depth, and height; and to know the love of Christ” (Ephesians 3:17–19, KJV). Come to Jesus, trust Him as your Savior, and follow Him as Lord.

Learning objectives

By the end of this lesson, you should be able to:

write complex numbers in Cartesian, polar and exponential forms;
convert between these forms by hand and with a calculator;
multiply, divide and raise complex numbers to integer powers in polar or exponential form;
interpret addition and subtraction as vector operations, and multiplication as rotation and scaling;
combine sinusoidal functions with the same frequency but different phase shifts.

1 · Cartesian, polar and exponential forms

In Lesson 1.12, a complex number was written in Cartesian form as $z=x+yi$ . Its modulus $r=|z|$ is its distance from the origin, and its argument $\theta$ is its angle from the positive real axis.

$r=\sqrt{x^2+y^2},\qquad x=r\cos\theta,\qquad y=r\sin\theta.$

Substituting the expressions for $x$ and $y$ gives the modulus–argument form, also called polar form.

$\boxed{z=r(\cos\theta+i\sin\theta) =r\,\mathrm{cis}\,\theta}$

Here $\mathrm{cis}\,\theta$ is shorthand for $\cos\theta+i\sin\theta$ . Euler’s formula provides a third representation, called exponential form.

$e^{i\theta}=\cos\theta+i\sin\theta \quad\Longrightarrow\quad \boxed{z=re^{i\theta}}.$

The forms describe the same complex number. For $z\ne0$ , $r>0$ ; the argument can be changed by any whole multiple of $2\pi$ without changing $z$ .

Angles: Exponential form normally uses radians. In this lesson, angles are in radians unless a degree symbol is shown. Set your calculator to radian mode for the examples using $\pi$ , $e^{i\theta}$ or a phase such as 10.

Worked example · Three representations

The complex number $z=1+\sqrt3\,i$ has modulus 2 and argument $\pi/3$ , because it lies in quadrant I.

$|z|=\sqrt{1^2+(\sqrt3)^2}=2, \qquad \arg(z)=\frac{\pi}{3}.$

$\boxed{1+\sqrt3\,i =2\left(\cos\frac{\pi}{3} +i\sin\frac{\pi}{3}\right) =2e^{i\pi/3}}.$

2 · Converting from Cartesian form

To convert $x+yi$ to polar form, calculate $r=\sqrt{x^2+y^2}$ and determine the argument from the location of $(x,y)$ on an Argand diagram. The signs of $x$ and $y$ tell you the quadrant.

Worked example · Check the quadrant

Convert $z=-\sqrt3+i$ to polar and exponential forms, using an argument between $0$ and $2\pi$ .

The point $(-\sqrt3,1)$ lies in quadrant II. Its modulus is 2, and its reference angle is $\pi/6$ .

$r=\sqrt{(-\sqrt3)^2+1^2}=2, \qquad\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}.$

$\boxed{z= 2\left(\cos\frac{5\pi}{6} +i\sin\frac{5\pi}{6}\right) =2e^{5\pi i/6}}.$

Calculator check: A calculator’s inverse tangent may return a reference angle rather than the argument in the correct quadrant. Check the signs of both Cartesian components.

3 · Converting back to Cartesian form

From $z=r(\cos\theta+i\sin\theta)$ , calculate the real part as $r\cos\theta$ and the imaginary part as $r\sin\theta$ .

Worked example · Polar to Cartesian

Write $4e^{-i\pi/6}$ in Cartesian form.

$4e^{-i\pi/6} =4\left(\cos\left(-\frac{\pi}{6}\right) +i\sin\left(-\frac{\pi}{6}\right)\right).$

$=\boxed{2\sqrt3-2i}.$

4 · Multiplication: stretch and rotate

When two complex numbers are multiplied, their moduli multiply and their arguments add.

$\boxed{ r_1e^{i\theta_1}\,r_2e^{i\theta_2} =r_1r_2e^{i(\theta_1+\theta_2)} }$

Geometrically, multiplying by $r_2e^{i\theta_2}$ stretches distances from the origin by $r_2$ and rotates points anticlockwise through $\theta_2$ .

Worked example · Product

Let $z_1=3e^{i\pi/6}$ and $z_2=2e^{i\pi/3}$ . Find their product in exponential and Cartesian forms.

$z_1z_2 =6e^{i(\pi/6+\pi/3)} =6e^{i\pi/2}.$

$\boxed{z_1z_2=6i}.$

5 · Division and integer powers

When dividing nonzero complex numbers in exponential form, divide their moduli and subtract their arguments.

$\boxed{ \frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}} =\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)} },\qquad r_2\ne0.$

For an integer power $n$ , raise the modulus to the power $n$ and multiply the argument by $n$ .

$\boxed{(re^{i\theta})^n=r^ne^{in\theta}}, \qquad n\in\mathbb Z.$

Worked example · Quotient

Find $\dfrac{6e^{5\pi i/6}}{2e^{-\pi i/3}}$ in exponential form. Give its principal argument between $-\pi$ and $\pi$ .

$\frac{6e^{5\pi i/6}}{2e^{-\pi i/3}} =3e^{i(5\pi/6+\pi/3)} =3e^{7\pi i/6}.$

The angle $7\pi/6$ is equivalent to $-5\pi/6$ , so:

$\boxed{z=3e^{-5\pi i/6}}.$

Worked example · An integer power

Find $(1+i)^8$ without expanding eight brackets.

Since $1+i=\sqrt2\,e^{i\pi/4}$ :

$(1+i)^8 =(\sqrt2)^8e^{i8\pi/4} =16e^{2\pi i} =\boxed{16}.$

Syllabus boundary: You need products, quotients and integer powers in polar or exponential form. Finding roots of complex numbers is not required in examinations for this syllabus section.

6 · Adding complex numbers as vectors

In an Argand diagram, $x+yi$ corresponds to the vector from the origin to $(x,y)$ . Addition combines its horizontal and vertical components; subtraction gives the vector difference.

Worked example · Vector interpretation

Let $z_1=2+i$ and $z_2=-1+3i$ .

$z_1+z_2=\boxed{1+4i}, \qquad z_1-z_2=\boxed{3-2i}.$

On the Argand diagram, the sum is the point $(1,4)$ . The difference is the point $(3,-2)$ .

Cartesian form is usually convenient for addition and subtraction; polar or exponential form is usually convenient for multiplication, division and powers.

7 · Combining sinusoidal functions

Two cosine functions with the same angular frequency can be combined into one cosine function. Represent each amplitude and phase by a complex number.

$A_1\cos(\omega t+\phi_1) +A_2\cos(\omega t+\phi_2) =A\cos(\omega t+B),$

where $A$ is the modulus and $B$ is the argument of the phasor sum:

$\boxed{Ae^{iB} =A_1e^{i\phi_1}+A_2e^{i\phi_2}}.$

Worked example · Two signals

Express $6\cos(5t)+10\cos(5t+\pi/3)$ as a single cosine function.

Add the phasors in Cartesian form.

$6+10e^{i\pi/3} =6+10\left(\frac12+\frac{\sqrt3}{2}i\right) =11+5\sqrt3\,i.$

Find its modulus and argument.

$A=\sqrt{11^2+(5\sqrt3)^2} =\sqrt{196} =14.$

$B=\tan^{-1}\left(\frac{5\sqrt3}{11}\right) \approx0.667.$

Therefore, to three significant figures:

$\boxed{6\cos(5t)+10\cos(5t+\pi/3) =14\cos(5t+0.667)}.$

TI-84 Plus CE: Use radian mode. For a phasor sum, calculate its real and imaginary components first. Then use $\sqrt{x^2+y^2}$ for the amplitude and determine the argument from the point’s quadrant.

IB-style Paper 1 · Skill practice

Question 1 · [5 marks]

Express $z=-3-3\sqrt3\,i$ in polar and exponential forms, using an argument between $-\pi$ and $\pi$ .

Show solution

The modulus is 6. The point lies in quadrant III, with reference angle $\pi/3$ . The principal argument is $-2\pi/3$ .

$|z|=\sqrt{(-3)^2+(-3\sqrt3)^2}=6.$

$\boxed{ z=6\left(\cos\left(-\frac{2\pi}{3}\right) +i\sin\left(-\frac{2\pi}{3}\right)\right) =6e^{-2\pi i/3} }.$

Question 2 · [4 marks]

Write $8e^{3\pi i/4}$ in Cartesian form.

Show solution

$8e^{3\pi i/4} =8\left(-\frac{\sqrt2}{2} +i\frac{\sqrt2}{2}\right) =\boxed{-4\sqrt2+4\sqrt2\,i}.$

Question 3 · [5 marks]

Let $z_1=4e^{i\pi/3}$ and $z_2=3e^{-i\pi/6}$ . Find $z_1z_2$ in exponential and Cartesian forms.

Show solution

$z_1z_2 =12e^{i(\pi/3-\pi/6)} =\boxed{12e^{i\pi/6}}.$

$12e^{i\pi/6} =12\left(\frac{\sqrt3}{2}+\frac12i\right) =\boxed{6\sqrt3+6i}.$

Question 4 · [5 marks]

Find $\dfrac{10e^{3\pi i/4}}{2e^{\pi i/4}}$ in exponential and Cartesian forms.

Show solution

$\frac{10e^{3\pi i/4}}{2e^{\pi i/4}} =5e^{i(3\pi/4-\pi/4)} =\boxed{5e^{i\pi/2}}.$

$\boxed{5e^{i\pi/2}=5i}.$

Question 5 · [6 marks]

Use polar form to find $(\sqrt3+i)^6$ exactly.

Show solution

The modulus of $\sqrt3+i$ is 2 and its argument is $\pi/6$ .

$\sqrt3+i=2e^{i\pi/6}.$

$(\sqrt3+i)^6 =2^6e^{i6\pi/6} =64e^{i\pi} =\boxed{-64}.$

Question 6 · [6 marks]

The number $z=2e^{i\pi/6}$ is multiplied by an unknown complex number $w$ . The result is $6e^{2\pi i/3}$ . Find $w$ in exponential and Cartesian forms.

Show solution

$w=\frac{6e^{2\pi i/3}}{2e^{i\pi/6}} =3e^{i(2\pi/3-\pi/6)} =\boxed{3e^{i\pi/2}}.$

$\boxed{w=3i}.$

Applications · More challenging IB-style questions

Application 1 · Voltage with different phase shifts [8 marks]

Two voltage sources are modeled, in volts, by:

$V_1=10\cos(40t),\qquad V_2=20\cos(40t+10),$

where $t$ is in seconds and angles are in radians. Express $V_1+V_2$ in the form $A\cos(40t+B)$ , giving $A$ and the principal value of $B$ to three significant figures.

Show solution

The phasor sum is $10+20e^{10i}$ . In Cartesian form:

$10+20e^{10i} =(10+20\cos10)+i(20\sin10) \approx-6.78143-10.88042i.$

It lies in quadrant III, so its principal argument is negative.

$A=\sqrt{(-6.78143)^2+(-10.88042)^2} \approx12.8207.$

$B=\arg(-6.78143-10.88042i) \approx-2.12815.$

$\boxed{V_1+V_2 \approx12.8\cos(40t-2.13)\ \mathrm{V}}.$

The angle $10$ is in radians; reducing it modulo $2\pi$ before calculating gives the same result.

Application 2 · Geometric transformation [8 marks]

Each point $z$ on an Argand diagram is transformed to $w=2e^{i\pi/3}z$ .

(a) Describe the geometric transformation. (b) Find the image of $z=1+i$ in Cartesian form. (c) Find the modulus of the image.

Show solution

(a) Multiplication by $2e^{i\pi/3}$ stretches distances from the origin by a factor of 2 and rotates anticlockwise through $\pi/3$ .

(b) Since $2e^{i\pi/3}=1+\sqrt3\,i$ :

$w=(1+\sqrt3\,i)(1+i) =\boxed{(1-\sqrt3)+(1+\sqrt3)i}.$

(c) The original number has modulus $\sqrt2$ , so the image has modulus:

$\boxed{|w|=2\sqrt2}.$

Application 3 · Cancellation and phase [8 marks]

Two signals with the same angular frequency are given by:

$S_1=12\cos(\omega t),\qquad S_2=12\cos\left(\omega t+\frac{2\pi}{3}\right).$

(a) Find the combined signal in the form $A\cos(\omega t+B)$ . (b) Explain why its amplitude is less than either individual amplitude.

Show solution

(a) Add the two phasors:

$12+12e^{2\pi i/3} =12+12\left(-\frac12+\frac{\sqrt3}{2}i\right) =6+6\sqrt3\,i.$

The resultant modulus is 12 and its argument is $\pi/3$ .

$\boxed{S_1+S_2 =12\cos\left(\omega t+\frac{\pi}{3}\right)}.$

(b) The combined amplitude is equal to either individual amplitude, not less. The proposed claim is false: the $120^\circ$ phase separation reduces the amplitude below their arithmetic sum of 24, but gives 12.