McIntyre MathsMathematics Extension 2
Extension 2 › Complex Numbers I › Lesson 5

Lesson 5 · Modulus–argument form

Mathematics Extension 2 · Complex Numbers I · Outcome ME2-12-03 · Term 4 · Week 8
Learning intentionTo represent complex numbers in modulus–argument (polar) form and convert between forms.
Success criteria · I can…
  • find ∣z∣|z| and arg⁡z\arg z, and the principal argument Arg⁡z∈(−π,π]\Arg z\in(-\pi,\pi]
  • write z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta), using a sketch to choose the quadrant
  • convert from polar form to Cartesian form with exact values
  • write z‾\conj{z} and −z-z in polar form

Lesson flow

LaunchWarm-up · Exact valuesSlide 2DefinitionsModulus and argumentSlide 3ExamplesExample 1 · Cartesian to polarSlide 4ExamplesExample 2 · Second and third quadrantsSlide 5ExamplesExample 3 · Choosing the quadrantSlide 6ExamplesExample 4 · On the axesSlide 7ExamplesExample 5 · Polar to CartesianSlide 8ExamplesExample 6 · Conjugate and negative in polar formSlide 9HSC-styleHSC practice · Polar formSlide 10ThinkSpot the error · ArgumentsSlide 11Exit ticketExit ticketSlide 12

Warm-up · Exact values

Slide 2
(a) tan⁡−13\tan^{-1}\sqrt3
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Answer
π3\dfrac{\pi}{3}
Why
tan⁡π3=3\tan\frac\pi3=\sqrt3, and π3\frac\pi3 lies in (−π2,π2)\left(-\frac\pi2,\frac\pi2\right).
(b) cos⁡2π3\cos\dfrac{2\pi}{3}
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Answer
−12-\dfrac12
Why
2π3\frac{2\pi}{3} is in the second quadrant, where cosine is negative, with reference angle π3\frac\pi3.
(c) sin⁡(−3π4)\sin\left(-\dfrac{3\pi}{4}\right)
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Answer
−12-\dfrac{1}{\sqrt2}
Why
−3π4-\frac{3\pi}{4} is in the third quadrant, where sine is negative, with reference angle π4\frac\pi4.
(d) Find θ∈(−π,π]\theta\in(-\pi,\pi] with cos⁡θ=−12\cos\theta=-\dfrac12 and sin⁡θ=−32\sin\theta=-\dfrac{\sqrt3}{2}.
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Answer
θ=−2π3\theta=-\dfrac{2\pi}{3}
Why
Both cosine and sine are negative, so θ\theta is in the third quadrant. The reference angle is π3\frac\pi3, so θ=−(π−π3)=−2π3\theta=-\left(\pi-\frac\pi3\right)=-\frac{2\pi}{3}.

Modulus and argument

Slide 3
Definition
Polar (modulus–argument) form
  • Modulus: r=∣z∣=x2+y2r=|z|=\sqrt{x^2+y^2}.
  • Argument: arg⁡z=θ\arg z=\theta, where cos⁡θ=xr\cos\theta=\frac xr and sin⁡θ=yr\sin\theta=\frac yr. It has many values, differing by multiples of 2π2\pi.
  • Principal argument: Arg⁡z\Arg z is the one value in (−π,π](-\pi,\pi]. The argument of 00 is undefined.
z=r(cos⁡θ+isin⁡θ)=r cis⁡θz=r(\cos\theta+i\sin\theta)=r\,\cis\theta
ReIm-6-5-4-3-2-1123456-6-5-4-3-2-11234560z = 3 + 4iθr = 5
r is the distance from the origin and θ is the angle measured anticlockwise from the positive real axis.
Watch out

Use radians in modulus–argument form. The principal argument lies in (−π,π](-\pi,\pi], so a negative real number has Arg⁡=π\Arg=\pi, not −π-\pi.

Discuss

(a) Sketch: mark rr and θ\theta for a point z=x+iyz=x+iy in each quadrant.
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Answer
In each quadrant, rr is the length of the line from OO to zz and θ\theta is the angle from the positive real axis to that line, anticlockwise positive and clockwise negative. For example, 3+4i3+4i has r=5r=5 and θ\theta marked above the axis, while 3−4i3-4i has the same rr and −θ-\theta below it.
Why
The sketch shows why the quadrant matters: cos⁡θ=xr\cos\theta=\frac xr and sin⁡θ=yr\sin\theta=\frac yr have signs that depend on where the point is.
(b) Write three different values of arg⁡(1+i)\arg(1+i). Which one is Arg⁡(1+i)\Arg(1+i)?
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Answer
For example π4\frac\pi4, 9π4\frac{9\pi}{4} and −7π4-\frac{7\pi}{4} (any π4+2kπ\frac\pi4+2k\pi). Arg⁡(1+i)=π4\Arg(1+i)=\frac\pi4.
Why
Adding or subtracting 2π2\pi gives the same point, so arg⁡\arg has many values. Only π4\frac\pi4 lies in (−π,π](-\pi,\pi].

Example 1 · Cartesian to polar

Slide 4
Plot z=3+4iz=3+4i. Find ∣z∣|z| and Arg⁡z\Arg z, and write zz in modulus–argument form. Repeat for z‾=3−4i\conj{z}=3-4i.
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Answer
∣z∣=5|z|=5 and Arg⁡z=tan⁡−143≈0.93\Arg z=\tan^{-1}\frac43\approx0.93, so z≈5(cos⁡0.93+isin⁡0.93)z\approx5(\cos0.93+i\sin0.93). Also z‾≈5(cos⁡(−0.93)+isin⁡(−0.93))\conj{z}\approx5(\cos(-0.93)+i\sin(-0.93)).
Why
r=32+42=5r=\sqrt{3^2+4^2}=5. The point is in the first quadrant, so θ=tan⁡−143≈0.93\theta=\tan^{-1}\frac43\approx0.93 radians (about 53∘53^\circ). The conjugate is the reflection in the real axis, so it has the same modulus and the opposite argument.

Example 2 · Second and third quadrants

Slide 5

Write in modulus–argument form using the principal argument. Sketch first!

(a) −1+i-1+i
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Answer
2(cos⁡3π4+isin⁡3π4)\sqrt2\left(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4}\right)
Why
r=2r=\sqrt2. The point is in the second quadrant with reference angle π4\frac\pi4, so θ=π−π4=3π4\theta=\pi-\frac\pi4=\frac{3\pi}{4}.
(b) −1−3 i-1-\sqrt3\,i
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Answer
2(cos⁡(−2π3)+isin⁡(−2π3))2\left(\cos\left(-\dfrac{2\pi}{3}\right)+i\sin\left(-\dfrac{2\pi}{3}\right)\right)
Why
r=1+3=2r=\sqrt{1+3}=2. The point is in the third quadrant with reference angle π3\frac\pi3, so θ=−(π−π3)=−2π3\theta=-\left(\pi-\frac\pi3\right)=-\frac{2\pi}{3}.
Watch out

A calculator gives tan⁡−1yx\tan^{-1}\frac yx in (−π2,π2)\left(-\frac\pi2,\frac\pi2\right). For (a) it gives −π4-\frac\pi4 and for (b) it gives π3\frac\pi3: it does not know the quadrant. A sketch fixes this.

ReIm-3-2-1123-3-2-11230−1 + i−1 − √3 i
Sketch the point first to choose the quadrant.

Example 3 · Choosing the quadrant

Slide 6
Key idea
Method
  1. Sketch the point.
  2. Find r=x2+y2r=\sqrt{x^2+y^2}.
  3. Find the acute angle α=tan⁡−1∣yx∣\alpha=\tan^{-1}\left|\frac yx\right|.
  4. Place θ\theta in the correct quadrant, with θ∈(−π,π]\theta\in(-\pi,\pi].
Exam practice(a) Find Arg⁡(6−2 i)\Arg\left(\sqrt6-\sqrt2\,i\right).
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Answer
−π6-\dfrac\pi6
Why
The point is in the fourth quadrant. tan⁡α=26=13\tan\alpha=\frac{\sqrt2}{\sqrt6}=\frac{1}{\sqrt3}, so α=π6\alpha=\frac\pi6 and Arg⁡=−π6\Arg=-\frac\pi6. (The modulus is 8=22\sqrt8=2\sqrt2.)
Exam practice(b) Write −4−3i-4-3i in polar form, with the argument correct to 2 decimal places.
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Answer
5(cos⁡(−2.50)+isin⁡(−2.50))5\left(\cos(-2.50)+i\sin(-2.50)\right)
Why
r=5r=5 (a 3-4-5 triangle). The point is in the third quadrant, with α=tan⁡−134≈0.64\alpha=\tan^{-1}\frac34\approx0.64. So Arg⁡=−π+0.64≈−2.50\Arg=-\pi+0.64\approx-2.50.

Example 4 · On the axes

Slide 7

Write in modulus–argument form using the principal argument.

(a) 5i5i
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Answer
5(cos⁡π2+isin⁡π2)5\left(\cos\dfrac\pi2+i\sin\dfrac\pi2\right)
Why
The point is on the positive imaginary axis, a quarter-turn from the real axis.
(b) −4-4
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Answer
4(cos⁡π+isin⁡π)4(\cos\pi+i\sin\pi)
Why
Arg⁡(−4)=π\Arg(-4)=\pi, not −π-\pi: the interval (−π,π](-\pi,\pi] includes π\pi and excludes −π-\pi.
(c) −i2-\dfrac{i}{2}
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Answer
12(cos⁡(−π2)+isin⁡(−π2))\dfrac12\left(\cos\left(-\dfrac\pi2\right)+i\sin\left(-\dfrac\pi2\right)\right)
Why
The point is on the negative imaginary axis, a quarter-turn clockwise from the real axis.

Example 5 · Polar to Cartesian

Slide 8

Express in the form x+iyx+iy, using exact values.

(a) 4(cos⁡π3+isin⁡π3)4\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)
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Answer
2+23 i2+2\sqrt3\,i
Why
4(12+32i)4\left(\frac12+\frac{\sqrt3}{2}i\right).
(b) 10 cis⁡2π310\,\cis\dfrac{2\pi}{3}
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Answer
−5+53 i-5+5\sqrt3\,i
Why
10(−12+32i)10\left(-\frac12+\frac{\sqrt3}{2}i\right).
(c) 6 cis⁡(−3π4)\sqrt6\,\cis\left(-\dfrac{3\pi}{4}\right)
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Answer
−3−3 i-\sqrt3-\sqrt3\,i
Why
6(−12−12i)=−3−3 i\sqrt6\left(-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}i\right)=-\sqrt3-\sqrt3\,i.
(d) 2 cis⁡π2\,\cis\pi
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Answer
−2-2
Why
2(−1+0i)=−22(-1+0i)=-2.

Example 6 · Conjugate and negative in polar form

Slide 9
Key idea
Geometry of z‾\conj{z} and −z-z

If z=r cis⁡θz=r\,\cis\theta, then z‾=r cis⁡(−θ)\conj{z}=r\,\cis(-\theta) (reflection in the real axis) and −z=r cis⁡(θ±π)-z=r\,\cis(\theta\pm\pi) (rotation of 180∘180^\circ). Choose +π+\pi or −π-\pi so that the argument lies in (−π,π](-\pi,\pi].

Let z=2 cis⁡π6z=2\,\cis\frac\pi6. Write in polar form:

(a) z‾\conj{z}
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Answer
2 cis⁡(−π6)2\,\cis\left(-\dfrac\pi6\right)
Why
Negate the argument.
(b) −z-z
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Answer
2 cis⁡(−5π6)2\,\cis\left(-\dfrac{5\pi}{6}\right)
Why
π6−π=−5π6\frac\pi6-\pi=-\frac{5\pi}{6}, which lies in (−π,π](-\pi,\pi].
(c) −z‾-\conj{z}
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Answer
2 cis⁡5π62\,\cis\dfrac{5\pi}{6}
Why
−π6+π=5π6-\frac\pi6+\pi=\frac{5\pi}{6}.
(d) Plot all four points.
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Answer
z=3+iz=\sqrt3+i, z‾=3−i\conj{z}=\sqrt3-i, −z=−3−i-z=-\sqrt3-i and −z‾=−3+i-\conj{z}=-\sqrt3+i: the vertices of a rectangle centred at the origin.
Why
All four have modulus 22, so they lie on a circle of radius 22.
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ReIm-3-2-1123-3-2-11230zz̄−z−z̄
The four points form a rectangle on the circle |z| = 2.

HSC practice · Polar form

Slide 10
Adapted from 2022 HSC Q11(c)(i)Let z=−3+iz=-\sqrt3+i. (i) Find ∣z∣|z| and Arg⁡z\Arg z. (ii) Write zz in the form r(cos⁡θ+isin⁡θ)r(\cos\theta+i\sin\theta).
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Answer
(i) ∣z∣=2|z|=2 and Arg⁡z=5π6\Arg z=\dfrac{5\pi}{6}. (ii) z=2(cos⁡5π6+isin⁡5π6)z=2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right).
Why
∣z∣=3+1=2|z|=\sqrt{3+1}=2. The point is in the second quadrant with tan⁡α=13\tan\alpha=\frac{1}{\sqrt3}, so α=π6\alpha=\frac\pi6 and Arg⁡z=π−π6=5π6\Arg z=\pi-\frac\pi6=\frac{5\pi}{6}.

Spot the error · Arguments

Slide 11

Each line is a student's answer. Find the error and correct it.

(a) z=−2+2iz=-2+2i: Arg⁡z=tan⁡−1(2−2)=−π4\Arg z=\tan^{-1}\left(\dfrac{2}{-2}\right)=-\dfrac\pi4
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Answer
Arg⁡z=3π4\Arg z=\dfrac{3\pi}{4}
Why
The point −2+2i-2+2i is in the second quadrant, but tan⁡−1\tan^{-1} only returns values in (−π2,π2)\left(-\frac\pi2,\frac\pi2\right). The reference angle is π4\frac\pi4, so Arg⁡z=π−π4=3π4\Arg z=\pi-\frac\pi4=\frac{3\pi}{4}. This is why the method says to sketch first.
(b) Arg⁡(−4)=−π\Arg(-4)=-\pi
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Answer
Arg⁡(−4)=π\Arg(-4)=\pi
Why
The principal argument lies in (−π,π](-\pi,\pi], which includes π\pi and excludes −π-\pi.
(c) 1−3 i=2(cos⁡π3+isin⁡π3)1-\sqrt3\,i=2\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)
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Answer
1−3 i=2(cos⁡(−π3)+isin⁡(−π3))1-\sqrt3\,i=2\left(\cos\left(-\dfrac\pi3\right)+i\sin\left(-\dfrac\pi3\right)\right)
Why
The point is in the fourth quadrant (positive real part, negative imaginary part), so the argument is −π3-\frac\pi3. The written form 2cis⁡π32\cis\frac\pi3 is the point 1+3 i1+\sqrt3\,i.
(d) 2(cos⁡π6−isin⁡π6)2\left(\cos\dfrac\pi6-i\sin\dfrac\pi6\right) has argument π6\dfrac\pi6
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Answer
The argument is −π6-\dfrac\pi6.
Why
The expression is not in polar form because of the minus sign. Since cosine is even and sine is odd, cos⁡π6−isin⁡π6=cos⁡(−π6)+isin⁡(−π6)\cos\frac\pi6-i\sin\frac\pi6=\cos\left(-\frac\pi6\right)+i\sin\left(-\frac\pi6\right).

Exit ticket

Slide 12
1. Write −23+2i-2\sqrt3+2i in polar form.
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Answer
4(cos⁡5π6+isin⁡5π6)4\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)
Why
r=12+4=4r=\sqrt{12+4}=4. The point is in the second quadrant, with α=tan⁡−1223=π6\alpha=\tan^{-1}\frac{2}{2\sqrt3}=\frac\pi6, so θ=5π6\theta=\frac{5\pi}{6}.
2. Express 6(cos⁡(−π4)+isin⁡(−π4))6\left(\cos\left(-\dfrac\pi4\right)+i\sin\left(-\dfrac\pi4\right)\right) as x+iyx+iy.
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Answer
32−32 i3\sqrt2-3\sqrt2\,i
Why
6(12−12i)=62−62i6\left(\frac{1}{\sqrt2}-\frac{1}{\sqrt2}i\right)=\frac{6}{\sqrt2}-\frac{6}{\sqrt2}i and 62=32\frac{6}{\sqrt2}=3\sqrt2.
3. Explain why Arg⁡(−5)=π\Arg(-5)=\pi and not −π-\pi.
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Answer
The principal argument lies in (−π,π](-\pi,\pi]: π\pi is included and −π-\pi is not.
Why
arg⁡(−5)\arg(-5) has the values π+2kπ\pi+2k\pi, including both π\pi and −π-\pi. Exactly one value must be chosen as Arg⁡\Arg, and the interval (−π,π](-\pi,\pi] is closed at π\pi and open at −π-\pi.