COMPLEX NUMBERS · HL
Polar Form
Modulus and argument.
Section 1 of 5
What polar form is
We usually write a complex number as $z = x + yi$. That's rectangular form.
Polar form describes the same number a different way: how far it is from the origin, and the angle it makes with the real axis.
(i) Building the formula
Plot $z = x + yi$ on an Argand diagram and drop a right-angled triangle to the real axis.
$r = \sqrt{x^2 + y^2}$ (distance from origin)
$\tan\theta = \dfrac{y}{x}$ (the angle)
From the same triangle:
$\cos\theta = \dfrac{x}{r} \;\Rightarrow\; x = r\cos\theta$
$\sin\theta = \dfrac{y}{r} \;\Rightarrow\; y = r\sin\theta$
Substitute both back into $z = x + yi$:
$z = r\cos\theta + i\,r\sin\theta$
$z = r(\cos\theta + i\sin\theta)$
$r$ is the modulus — written $|z|$ or mod$(z)$. $\theta$ is the argument — written arg$(z)$.
(ii) Worked example — write $1 + i$ in polar form
Here $x = 1$ and $y = 1$.
$r = \sqrt{1^2 + 1^2} = \sqrt{2}$
$\tan\theta = \dfrac{1}{1} = 1 \;\Rightarrow\; \theta = 45^{\circ}$
$1 + i = \sqrt{2}\,(\cos 45^{\circ} + i\sin 45^{\circ})$
Must learn
1.$z = r(\cos\theta + i\sin\theta)$
2.$r = |z| = \sqrt{x^2 + y^2}$ and $\tan\theta = \dfrac{y}{x}$
YOU TRY · 1
Write $\sqrt{3} + i$ in polar form.
First quadrant. Find $r$, then $\theta$ from $\tan\theta$.
$r = \sqrt{3 + 1} = 2$, $\tan\theta = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = 30^{\circ}$
$2(\cos 30^{\circ} + i\sin 30^{\circ})$
YOU TRY · 2
Write $2 + 2i$ in polar form.
Still first quadrant.
$r = \sqrt{4 + 4} = 2\sqrt{2}$, $\tan\theta = 1 \Rightarrow \theta = 45^{\circ}$
$2\sqrt{2}\,(\cos 45^{\circ} + i\sin 45^{\circ})$
Section 2 of 5
The angle in any quadrant
When $z$ is not in the first quadrant, $\tan\theta = \dfrac{y}{x}$ on its own will mislead you. Find the reference angle $\alpha$ first, then adjust for the quadrant.
$\alpha$ is the acute angle the line makes with the real axis: $\tan\alpha = \dfrac{|y|}{|x|}$.
Quadrant rule
Q1.$\theta = \alpha$
Q2.$\theta = 180^{\circ} - \alpha$
Q3.$\theta = 180^{\circ} + \alpha$
Q4.$\theta = 360^{\circ} - \alpha$
(i) $1 + \sqrt{3}\,i$ (Q1)
$x = 1, \quad y = \sqrt{3}$
$r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$
$\tan\theta = \sqrt{3} \;\Rightarrow\; \theta = 60^{\circ}$
$1 + \sqrt{3}\,i = 2(\cos 60^{\circ} + i\sin 60^{\circ})$
(ii) $-1 - i$ (Q3)
$x = -1, \quad y = -1$
$r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$
$\tan\alpha = 1 \;\Rightarrow\; \alpha = 45^{\circ}$
Third quadrant: $\theta = 180^{\circ} + 45^{\circ} = 225^{\circ}$
$-1 - i = \sqrt{2}\,(\cos 225^{\circ} + i\sin 225^{\circ})$
(iii) $-2 + 2\sqrt{3}\,i$ (Q2)
$x = -2, \quad y = 2\sqrt{3}$
$r = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$
$\tan\alpha = \dfrac{2\sqrt{3}}{2} = \sqrt{3} \;\Rightarrow\; \alpha = 60^{\circ}$
Second quadrant: $\theta = 180^{\circ} - 60^{\circ} = 120^{\circ}$
$-2 + 2\sqrt{3}\,i = 4(\cos 120^{\circ} + i\sin 120^{\circ})$
(iv) $2 - 2\sqrt{3}\,i$ (Q4)
Same magnitudes as above, so $r = 4$.
Fourth quadrant: $\theta = 360^{\circ} - 60^{\circ} = 300^{\circ}$
$2 - 2\sqrt{3}\,i = 4(\cos 300^{\circ} + i\sin 300^{\circ})$
YOU TRY · 3
Write $-2 - 2i$ in polar form.
Which quadrant? Find $\alpha$ first.
Q3. $r = 2\sqrt{2}$, $\tan\alpha = 1 \Rightarrow \alpha = 45^{\circ}$, $\theta = 180^{\circ} + 45^{\circ} = 225^{\circ}$
$2\sqrt{2}\,(\cos 225^{\circ} + i\sin 225^{\circ})$
YOU TRY · 4
Write $-\sqrt{3} + i$ in polar form.
Negative real part, positive imaginary part.
Q2. $r = 2$, $\tan\alpha = \dfrac{1}{\sqrt{3}} \Rightarrow \alpha = 30^{\circ}$, $\theta = 180^{\circ} - 30^{\circ} = 150^{\circ}$
$2(\cos 150^{\circ} + i\sin 150^{\circ})$
Section 3 of 5
Multiply, divide, invert
Polar form makes multiplication and division easy. Take two numbers:
$z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$
$z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$
(i) Product
Multiply the moduli, add the arguments.
$z_1 z_2 = r_1 r_2\big(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\big)$
(ii) Quotient
Divide the moduli, subtract the arguments.
$\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\big(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\big)$
(iii) Reciprocal
$\dfrac{1}{z_1} = \dfrac{1}{r_1}\big(\cos\theta_1 - i\sin\theta_1\big)$
The three properties
1.Product: multiply moduli, add arguments.
2.Quotient: divide moduli, subtract arguments.
3.Reciprocal: $\dfrac{1}{r}$, flip the sign on the $\sin$.
YOU TRY · 5
If $z_1 = 3(\cos 20^{\circ} + i\sin 20^{\circ})$ and $z_2 = 2(\cos 50^{\circ} + i\sin 50^{\circ})$, write $z_1 z_2$ in polar form.
Multiply the moduli, add the angles.
$r = 3 \times 2 = 6$, $\theta = 20^{\circ} + 50^{\circ} = 70^{\circ}$
$6(\cos 70^{\circ} + i\sin 70^{\circ})$
Section 4 of 5
Back to rectangular form
Rectangular form is $x + yi$. So "find in rectangular form" means combine using the property, then evaluate the $\cos$ and $\sin$ and tidy up.
(i) Worked example
$z_1 = 4(\cos 30^{\circ} + i\sin 30^{\circ}), \quad z_2 = 8(\cos 60^{\circ} + i\sin 60^{\circ})$
Find $z_1 z_2$ in rectangular form.
Multiply moduli, add arguments: $z_1 z_2 = 4 \times 8\,(\cos 90^{\circ} + i\sin 90^{\circ})$
$= 32(0 + i)$
$z_1 z_2 = 32i$
Now find $\dfrac{z_2}{z_1}$ in rectangular form.
Divide moduli, subtract arguments: $\dfrac{z_2}{z_1} = \dfrac{8}{4}\,(\cos 30^{\circ} + i\sin 30^{\circ})$
$= 2\left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i\right)$
$\dfrac{z_2}{z_1} = \sqrt{3} + i$
(ii) Radians work the same way
Angles can be in radians. Recall $\pi = 180^{\circ}$.
$z = 5\left(\cos \dfrac{\pi}{6} + i\sin \dfrac{\pi}{6}\right), \quad w = 8\left(\cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3}\right)$
Find $zw$.
$zw = 40\left(\cos\left(\dfrac{\pi}{6} + \dfrac{\pi}{3}\right) + i\sin\left(\dfrac{\pi}{6} + \dfrac{\pi}{3}\right)\right)$
$= 40\left(\cos \dfrac{\pi}{2} + i\sin \dfrac{\pi}{2}\right) = 40(\cos 90^{\circ} + i\sin 90^{\circ})$
$zw = 40i$
YOU TRY · 6
If $z_1 = 2(\cos 40^{\circ} + i\sin 40^{\circ})$ and $z_2 = 3(\cos 50^{\circ} + i\sin 50^{\circ})$, find $z_1 z_2$ in rectangular form.
Combine in polar first, then evaluate.
$6(\cos 90^{\circ} + i\sin 90^{\circ}) = 6(0 + i)$
$6i$
Section 5 of 5
Where they come from
(i) Prove the product rule
Multiply the two numbers out, remembering $i^2 = -1$:
$z_1 z_2 = r_1(\cos\theta_1 + i\sin\theta_1)\cdot r_2(\cos\theta_2 + i\sin\theta_2)$
$= r_1 r_2\big(\cos\theta_1\cos\theta_2 + i^2\sin\theta_1\sin\theta_2 + i\cos\theta_1\sin\theta_2 + i\sin\theta_1\cos\theta_2\big)$
$= r_1 r_2\big(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2 + i(\cos\theta_1\sin\theta_2 + \sin\theta_1\cos\theta_2)\big)$
Now apply the compound-angle formulae:
Compound angles
1.$\cos(A + B) = \cos A\cos B - \sin A\sin B$
2.$\sin(A + B) = \sin A\cos B + \cos A\sin B$
$z_1 z_2 = r_1 r_2\big(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\big)$
(ii) Prove the quotient rule
Multiply top and bottom by the conjugate of the denominator, $\cos\theta_2 - i\sin\theta_2$:
$\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\cdot\dfrac{(\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 - i\sin\theta_2)}{(\cos\theta_2 + i\sin\theta_2)(\cos\theta_2 - i\sin\theta_2)}$
Denominator: $\cos^2\theta_2 + \sin^2\theta_2 = 1$
Numerator: $\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2 + i(\sin\theta_1\cos\theta_2 - \cos\theta_1\sin\theta_2)$
These are the compound-angle formulae for $(\theta_1 - \theta_2)$.
$\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\big(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\big)$
SUM
The lot in one box
Polar form toolkit
1.$z = r(\cos\theta + i\sin\theta)$, with $r = |z| = \sqrt{x^2 + y^2}$, $\tan\theta = \dfrac{y}{x}$.
2.Reference angle $\alpha$ first, then quadrant: Q1 $\theta = \alpha$; Q2 $180^{\circ} - \alpha$; Q3 $180^{\circ} + \alpha$; Q4 $360^{\circ} - \alpha$.
3.Product: multiply moduli, add arguments.
4.Quotient: divide moduli, subtract arguments.
5.Reciprocal: $\dfrac{1}{z} = \dfrac{1}{r}(\cos\theta - i\sin\theta)$.
6.Rectangular form $= x + yi$.
End of lesson
Polar Form — HL · Mathslive.ie