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Wize University Linear Algebra Textbook > Complex Numbers

Polar Form

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Polar Form

Consider a complex number in standard form: z=a+biz=a+bi.
Polar form is an another way of representing complex numbers, where we define:
  • Magnitude: r=∣z∣=a2+b2\boxed{r = |z| = \sqrt{a^2 + b^2}}
  • Direction: θ\theta is the angle swept out (in radians) from the positive x-axis to the vector pointing to z=a+biz=a+bi.
  • real component (x-component): a=rcos⁡θa=r\cos\theta
  • imaginary component (y-component): b=rsin⁡θb=r\sin\theta
  • tan⁡(θ)=ba  ⟹  θ=tan⁡−1(ba)\tan(\theta)=\dfrac{b}{a} \implies \boxed{\theta=\tan^{-1}\left(\dfrac{b}{a}\right)}

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Polar Form

Using trigonometry, z=a+biz=a+bi can be written as z=rcos⁡θ⏟a+rsin⁡θ⏟b⋅iz=\underbrace{r\cos\theta}_{a} + \underbrace{r\sin\theta}_{b} \cdot i.
Simplifying, we obtain the polar form of zz:
z=r(cos⁡θ+isin⁡θ)\boxed{\quad z = r\big(\cos\theta + i\sin\theta\big) \quad}
We sometimes use the more compact notation, cis(θ)=cos⁡(θ)+isin⁡(θ)\textcolor{blue}{\text{c}} \textcolor{green}{\text{i}} \textcolor{red}{\text{s}}(\theta) = \textcolor{blue}{\cos}(\theta) +\textcolor{green}{i} \textcolor{red}{\sin}(\theta):
z=r cis(θ)\boxed{\quad z=r\,\text{cis}(\theta) \quad}
Notes:
  • It is standard convention for θ\theta to be in the interval (−π,π](-\pi,\pi].
  • θ\theta is also called the argument of zz, denoted arg(z)\text{arg}(z).
Wize Tip
The range of tan⁡−1\tan^{-1} is only (−π2,π2)\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right) (bottom-right and top-right quadrants), so take care when finding θ\theta:
  • If zz is in the top-left quadrant, add π\pi to tan⁡−1(ba)\tan^{-1}\left(\dfrac{b}{a}\right)
  • If zz is in the bottom-left quadrant, subtract π\pi to tan⁡−1(ba)\tan^{-1}\left(\dfrac{b}{a}\right)

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Euler's Formula

Euler's formula states:
cos⁡(θ)+isin⁡(θ)=eiθ\boxed{\quad \cos(\theta)+i\sin(\theta) = e^{i\theta} \quad}
Therefore, every complex number can be written as:
z  =  r(cos⁡(θ)+isin⁡(θ))  =  reiθz \;=\; r \big( \cos(\theta) + i\sin(\theta)\big) \;=\; \boxed{re^{i\theta}}

The expression reiθre^{i\theta} is sometimes called the exponential form of zz, but we will refer to this as the polar form as well.
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Multiplication

Multiplying complex numbers is much faster in polar form.
Let z=reiθz=re^{i\theta}, w=seiϕw=se^{i\phi} be complex numbers. Then:
z⋅w=(reiθ)⋅(seiϕ)=(rs)ei(θ + ϕ)z\cdot w = (\colorOne{r}e^{i\colorFive{\theta}}) \cdot (\colorOne{s}e^{i\colorFive{\phi}}) = (\colorOne{rs})e^{i(\colorFive{\theta\,+\,\phi})}
Wize Tip
To multiply two complex numbers: multiply the magnitudes, and add the angles.

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Geometric Interpretation of Multiplication

Multiplying by ii is effectively a rotation by 90 degrees (counter-clockwise).
Why?
Consider the polar form of the complex number z=i    ⟹    z=0+1iz=i \ \ \implies\ \ z=0+1i:
r=02+12=1θ=π2(since i is on the positive y-axis)  ⟹  z=i=eiπ/2\begin{aligned} r &= \sqrt{0^2 + 1^2} = \bm1\\ \theta &= \dfrac{\bm\pi}{\bm2} \quad \text{(since $i$ is on the positive $y$-axis)}\\[0.5em] \implies z &= \boxed{i=e^{i \pi/2}} \end{aligned}
Now consider what happens when we multiply any complex number w=seiϕw=se^{i\phi} in polar form by z=i=eiπ/2z=i=e^{i\pi/2}:
w⋅z=seiϕ⋅eiπ/2=sei(ϕ + π/2)\begin{aligned} w \cdot z &= s e^{i \colorFive\phi} \cdot e^{i \colorFive{\pi/2}}\\ &= s e^{i\colorFive{(\phi \ +\ \pi/2)}} \end{aligned}
Wize Concept
Recall, multiplying complex numbers in polar form is easy: multiply the magnitudes, and add the angles
The result has the same magnitude, but we added π/2\pi/2 to the existing angle: this is a rotation by 90 degrees!
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Example
Consider the complex number w = 1+3iw \ =\ 1+\sqrt{3} i. In polar form, we get w = 2eiπ/3 = 2ei2π/6w \ =\ 2e^{i \pi/3} \ =\ 2e^{i2\pi/6}.
Repeated multiplication by ii is the same as adding π2=3π6\dfrac{\pi}{2} = \dfrac{3\pi}{6} to the angle each time:

However, notice that the angle 14π/6=π/3+2π14\pi/6 = \pi/3 + 2\pi, so we have come full circle. We rotated 90 degrees four times!


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Example: Polar Form of Complex Numbers

Part A)

Write the complex number z=−1+3iz=-1+\sqrt{3}i in polar form (both ways).
Considering the standard from is written z=a+biz=a+bi , we see that a=−1a=-1 and b=3b=\sqrt{3}.
r=a2+b2=(−1)2+(3)2=2r=\sqrt{a^2+b^2}=\sqrt{(-1)^2+(\sqrt{3})^2}=2
To find θ\theta we note that zz is in quadrant 2 (top-left) of the complex plane, so we must add π\pi to the result of tan⁡−1(ba)\tan^{-1}\left( \dfrac{b}{a} \right):
θ=tan⁡−1(3−1)+π=2π3\theta=\tan^{-1}\left(\dfrac{\sqrt{3}}{-1}\right)+\pi=\dfrac{2\pi}{3}
Then the polar form of zz is z=2(cos⁡(2π3)+isin⁡(2π3))  =  2ei2π/3\boxed{ z= 2 \left( \cos\left( \frac{2\pi}{3} \right)+i\sin \left (\frac{2\pi}{3} \right) \right)\;=\;2e^{i2\pi/3} }

Part B)

Write the complex number w=3(cos⁡(−π2)+isin⁡(−π2))w=\sqrt{3}\Big(\cos(-\frac{\pi}{2})+i\sin(-\frac{\pi}{2})\Big) in standard form.
w=3(cos⁡(−π/2)+isin⁡(−π/2))=3(0+i(−1))=−3i\begin{aligned} w &=\sqrt{3}\left(\cos(-{\pi}/{2})+i\sin(-{\pi}/{2})\right)\\[0.5em] &=\sqrt{3}\Big(0+i(-1)\Big)\\[0.5em] &=\boxed{-\sqrt{3}i}\\[0.5em] \end{aligned}

Part C)

Write the complex number u=4e−iπ/6u=4e^{-i\pi/6} in standard form.
u=4e−iπ/6=4(cos⁡(−π/6)+isin⁡(−π/6))=4(32+i(−12))=23−2i\begin{aligned} u &=4e^{-i\pi/6}\\ &=4\left(\cos(-{\pi}/{6})+i\sin(-{\pi}/{6})\right)\\[0.5em] &=4\left(\dfrac{\sqrt3}{2}+i\left( -\dfrac{1}{2} \right)\right)\\[0.5em] &= \boxed{2\sqrt{3}-2i} \end{aligned}

Part D)

Multiply z=−1+3iz=-1+\sqrt{3}i and u=4e−iπ/6u=4e^{-i\pi/6}, and write the product in standard form.
We can use the polar form of zz to simplify multiplication.
z=−1+3i  ⟹  r=(−1)2+(3)2=2θ=tan⁡−1(ba)+π=2π3  ⟹  z=2ei2π/3\begin{array}{rclllll} z &=& -1+\sqrt{3}i\\[1em] \implies r &=& \sqrt{(-1)^2 + (\sqrt3)^2} &=& 2\\[0.5em] \theta &=& \tan^{-1}\left( \dfrac{b}{a} \right) \colorTwo{+ \pi} &=& \dfrac{2\pi}{3}\\[1em] \implies z &=& 2 e^{i2\pi/3} \end{array}
Now we multiply:
z⋅u=2ei2π/3⋅4e−iπ/6=(2⋅4)ei((2π/3)−(π/6))=8eiπ/2=8(cos⁡(π2)+isin⁡(π2))=8(0+i(1))=8i\begin{array}{rcl} z\cdot u &=& 2e^{i2\pi/3}\cdot4e^{-i\pi/6}\\[0.5em] &=& (2\cdot 4)e^{i((2\pi/3)-(\pi/6))}\\[0.5em] &=& 8e^{i\pi/2}\\[0.5em] &=& 8 \left(\cos(\frac{\pi}{2}) +i\sin(\frac{\pi}{2}) \right)\\[0.5em] &=& 8 \left(0 +i(1) \right)\\[0.5em] &=& \boxed{8i} \end{array}

Practice: Polar Form

Write z=1−iz=1-i in polar form.

Practice: Multiplying Complex Numbers

Using polar form, find the product of z=3−33iz=3-3\sqrt{3}i and w=cis(π6)w=\text{cis}\left(\dfrac{\pi}{6}\right), and write the answer in standard form.