Polar Coordinates

We can place a point in a plane by polar coordinates \((r, \ \theta).\) Euler's formula states that \(e^{i\theta} = \cos{\theta} + i \sin{\theta}.\) Employing this formula, we have \[r e^{i\theta} = r \cos{\theta} + i r \sin{\theta} = x + iy,\] so we have Cartesian coordinates \((x, \ y)\) from the polar coordinates. Therefore, a point in a plane can be represented by polar coordinates \((r, \ \theta),\) which have their equivalent complex numbers \(r e^{i\theta}.\)

The multiplication of two points \((r_1, \ \theta_1)\) and \((r_2, \ \theta_2),\) which have their equivalent complex numbers, can be easily calculated. Given two equivalent complex numbers \(r_1 e^{i\theta_1}\) and \(r_2 e^{i\theta_2},\) the multiplication is \[\begin{align} re^{i\theta} =& \left(r_1 e^{i \theta _1}\right)\left(r_2 e^{i \theta _2}\right) \\ =& (r_1 r_2)\left( e^{i\theta _1} e^{i\theta _2}\right)\\ =& r_1 r_2 e^{i (\theta _1 + \theta _2)}, \end{align}\] where the rule of addition of exponents over real numbers is extended to that over complex numbers in the same way. Therefore, the multiplication of two points is \((r, \ \theta),\) where \(r = r_1 r_2\) and \(\theta = \theta _1 + \theta _2.\)

What is the multiplication of two points \((3, 5)\) and \((2, 4)\) in polar coordinates?

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The first coordinate of the multiplication is the product of the two first coordinates. The second coordinate of the multiplication is the sum of the two second coordinates. Therefore, we have \[(r, \theta) = (3 \times 2, 5 + 4) = (6, 9). \ _\square \]

What is the multiplication of two points \((5, 2)\) in polar coordinates and \((4, 3)\) in Cartesian coordinates? Approximate the angles in radians up to two digits below the decimal point.

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Convert \((4, 3)\) in Cartesian coordinates to a representation in polar coordinates: \(\left(\sqrt{4^2 + 3^2}, \arctan{\frac{3}{4}}\right) = (5, 0.64).\)

In polar coordinates, the first coordinate of the multiplication is the product of the two first coordinates, and the second coordinate of the multiplication is the sum of the two second coordinates. Therefore, we have \[(r, \theta) \approx (5 \times 5, 2 +0.64) = (25, 2.64). \ _\square \]

What is the multiplication of two points \((5, -12)\) and \((6, 8)\) in Cartesian coordinates? Approximate the angles in radians up to two digits below the decimal point.

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Convert \((5, -12)\) and \((6, 8)\) in Cartesian coordinates to representations in polar coordinates: \[\left(\sqrt{5^2 + (-12)^2}, \arctan{\frac{-12}{5}}\right) = (13, -1.176)\] and \[\left(\sqrt{6^2 + 8^2}, \arctan{\frac{8}{6}}\right) = (10, 0.927 ).\]

Therefore, we have \[(r, \theta) = (13 \times 10, -1.176 \ldots + 0.927 \ldots) \approx (130, -0.25). \ _\square \]

Main article: Polar to cartesian, Cartesian to Polar

A complex number \(z = x + \imath y \) is represented by the point \((x,y)\) in the complex plane. From the properties of complex numbers, we can write \[ \begin{align} x &= \Re(z) = |z| \cos(\theta)\\ y &= \Im(z) = |z| \sin(\theta), \end{align} \] where \(|z| = \sqrt{x^2 + y^2}\). This is shown in the picture below:

complex numbers

Note that the \((x,y)\) pair can equivalently be described by trigonometric functions of another pair \((|z|, \theta)\), often denoted \((r, \theta)\). These are referred to as the polar coordinates of the complex number \(z\).

\(r\) is a non-negative number denoting the magnitude of the complex number (the radius of the circle) and is represented on the radial axis that extends outward from the origin at \((0,0)\). An expression for \(\theta\) can be obtained by dividing \(y = |z| \sin (\theta) \) by \(x = |z| \cos (\theta)\):

\[ \theta = \arctan\left(\frac{y}{x}\right) \] and is called the argument of the complex number.

What is the polar form of the complex number \(z = 1 + 1i?\)

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We can draw this point on the complex plane:

example

We can see from the image that the triangle has two equal sides; from this we know that it must also have two equal angles. Since one of the angles is \(90\) degrees, we can safely conclude that \(\theta\) is equal to \(45\) degrees or \(\frac{\pi}{4}\) radians. We can then find \(|z|\) by computing the hypotenuse of the triangle using the Pythagoras theorem: \[x^2 + y^2 = z^2 \implies |z| = \sqrt{1^2 + 1^2} = \sqrt{2}. \] Thus we can write our complex number as \[z = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right ).\ _\square\]

Write \( 1 + \sqrt{3} i \) in its polar form.

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Applying our formula, \[\begin{align} \theta &= \arg(z)\\ &= \arctan \frac{y}{x}\\ &= \arctan \sqrt{3}= \frac{\pi}{3} \\\\ r &= |z| \\&= \sqrt{ 1^2 + (\sqrt{3}) ^2} = 2\\\\ z &= |z|(\cos \theta + i \sin \theta ) \\ z &= 2\left(\cos \frac{\pi}{3} + i \sin\frac{\pi}{3} \right).\ _\square \end{align}\]

Write \(6\left(\cos \frac{5\pi}{3} +i \sin \frac{5\pi}{3} \right)\) in its rectangular form.

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We know that \(|z| = 6,\) and we evaluate \[\sin \frac{5\pi}{3} = \frac{-\sqrt{3}}{2}, \quad \cos \frac{5\pi}{3} = \frac{1}{2}. \] Thus our complex number \(z\) is \[ \begin{align} z &= |z|(\cos \theta + i\sin \theta )\\ &= 6\left( \frac{1}{2} + \frac{-\sqrt{3}}{2} i\right)\\ &= 3\left(1 - \sqrt{3} i\right) \\ &= 3 -3\sqrt{3}i.\ _\square \end{align}\]

\( \left(1 + \sqrt{2} i\right)\cdot\left(1-\sqrt{2} i\right) \) in its polar form.

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We have \[\begin{align} (1 + \sqrt{2} i)\cdot(1-\sqrt{2} i) &= 1\cdot1 - \sqrt{2}i +\sqrt{2}i - \sqrt{2}i \cdot \sqrt{2}i \\ &= 3 + 0\cdot i. \end{align}\] Then from our formula, we have \[\theta = arg(z)= \arctan \frac{y}{x}= \arctan \frac{0}{3}= 0.\] Then since \( r=\lvert z \rvert = \sqrt{3^2 +0^2 } = 3,\) we can conclude that \[\begin{align} z &= |z|(\cos \theta + i \sin \theta ) \\ &= 3(\cos 0 + i \sin 0 ).\ _\square \end{align}\]

Write \( i\left(1-\sqrt{3}i \right) \) in its polar form.

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The given expression can be rewritten as \[\begin{align} i(1-\sqrt{3}i ) = \sqrt{3}+i. \end{align}\] Then we have \[\theta = arg(z)= \arctan \frac{y}{x}= \arctan \frac{1}{\sqrt{3}}= \frac{\pi}{6},\] and \[r = |z| = \sqrt{ 1^2 + (\sqrt{3}) ^2} = 2.\] Thus, the polar form \(z\) is \[\begin{align} z &= |z|(\cos \theta + i \sin \theta ) \\ &= 2\left(\cos \frac{\pi}{6} + i \sin\frac{\pi}{6} \right).\ _\square \end{align}\]

If \( b \left ( \cos \frac{\pi}{a} + \sin \frac{\pi}{a}i \right ) \) is the polar form of \( 3+ 3\sqrt{3}i ,\) what are \(a\) and \(b?\)

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Observe that \( 3+ 3\sqrt{3} i \) can be expressed as \( c \left ( \frac{3}{c} + \frac{3\sqrt{3}}{c} i \right ),\) where \( c \) is a positive integer. Then since \( \sin^2 \theta + cos^2 \theta =1,\) we can get the value of \(c\) as follows: \[\begin{align} \left ( \frac{3}{c} \right )^2 + \left ( \frac{3\sqrt{3}}{c} \right )^2 = \frac{ 3^2 + (3\sqrt{3})^2}{c^2} = \frac{36}{c^2} &=1 \\ \Rightarrow c&=6.
\end{align}\] Since \(c=6,\) \( 3+ 3\sqrt{3} i \) can be expressed as \[c\left( \frac{3}{c} + \frac{3\sqrt{3}}{c} i \right ) = 6\left ( \frac{3}{6} + \frac{3\sqrt{3}}{6}i \right ) = 6\left ( \frac{1}{2} + \frac{\sqrt{3}}{2}i\right ).\] Thus, since \( \cos \frac{\pi}{a} = \frac{1}{2}\) and \( \sin \frac{\pi}{a} = \frac{\sqrt{3}}{2},\) from trigonometry we can conclude that \(a=3.\) Therefore, the answer is \[ a=3, b=6. \ _\square \]