3.2 Polar form

Example 1

Given \displaystyle z=2+i and \displaystyle w=-3-i. Indicate \displaystyle z, \displaystyle w, \displaystyle \overline{z}, \displaystyle \overline{z}-\overline{w} and \displaystyle z-w in the complex plane.

Note that complex conjugate pairs are mirror images in the real axis.

Example 2

Indicate in the complex plane all numbers \displaystyle z that meet the following conditions:

1. \displaystyle \mathop{\rm Re} z \ge 3\,,
2. \displaystyle -1 < \mathop{\rm Im} z \le 2\,.

The first inequality defines the region in the figure on the left below, and the second inequality defines the region in the figure on the right below.

Example 3

Indicate the following sets in the complex plane:

Example 4

Indicate in the complex plane all numbers \displaystyle z satisfying the following conditions:

1. \displaystyle \, \left\{ \eqalign{&|\,z-2i\,|\le 3\cr &1\le\mathop{\rm Re} z\le 2}\right.
   
   The first inequality gives the points on and inside a circle with radius 3 and center at \displaystyle 2i. The second inequality is a vertical strip of points with their real part between 1 and 2. The area satisfying both inequalities is given by the points which lie both within the circle and within the strip.




2. \displaystyle \, |\,z+1\,|=|\,z-2\,|
   
   The equation can be written as \displaystyle |\,z-(-1)\,|=|\,z-2\,|. This shows then that \displaystyle z should be at an equal distance from \displaystyle -1 and \displaystyle 2. This condition is met by all the numbers \displaystyle z that have a real part \displaystyle 1/2.

Example 5

Write the following complex numbers in polar form:

1. \displaystyle \,\,-3
   
   We have that \displaystyle |\,-3\,|=3 and \displaystyle \arg (-3)=\pi, which means that \displaystyle \ -3=3(\cos\pi+i\,\sin\pi).
2. \displaystyle \,i
   
   We have that \displaystyle |\,i\,|=1 and \displaystyle \arg i = \pi/2 which in polar form is \displaystyle \ i=\cos(\pi/2)+i\,\sin(\pi/2)\,.
3. \displaystyle \,1-i
   
   The formula for a the absolute value of a complex number gives \displaystyle |\,1-i\,|=\sqrt{1^2+(-1)^2}=\sqrt{2}. The complex number lies in the fourth quadrant and has an angle \displaystyle \pi/4 with the positive real axis, which gives \displaystyle \arg (1-i)=2\pi-\pi/4=7\pi/4. Thus \displaystyle \ 1-i=\sqrt{2}\,\bigl(\cos(7\pi/4)+i\sin(7\pi/4)\,\bigr).
4. \displaystyle \,2\sqrt{3}+2i
   
   The absolute value is the easiest to calculate

   \displaystyle |\,2\sqrt{3}+2i\,|=\sqrt{(2\sqrt{3}\,)^2+2^2}=\sqrt{16}=4\,\mbox{.}

   If we call the argument \displaystyle \alpha then it satisfies the relationship

   \displaystyle \tan\alpha=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}

   and since the number is in the first quadrant (positive real and imaginary parts) one gets \displaystyle \alpha=\pi/6 and we have that

   \displaystyle 2\sqrt{3}+2i=4\bigl(\cos\frac{\pi}{6}+i\,\sin\frac{\pi}{6}\bigr)\,\mbox{.}

Example 6

Simplify the following expressions by writing them in polar form:

1. \displaystyle \Bigl(\frac{1}{\sqrt2} -\frac{i}{\sqrt2}\Bigr) \Big/ \Bigl( -\frac{1}{\sqrt2} +\frac{i}{\sqrt2}\Bigr)
   
   We write the numerator and denominator in polar form

   \displaystyle \begin{align*}\frac{1}{\sqrt2} -\frac{i}{\sqrt2} &= 1\times\Bigl(\cos\frac{7\pi}{4}+i\,\sin\frac{7\pi}{4}\Bigr)\\[4pt] -\frac{1}{\sqrt2} +\frac{i}{\sqrt2} &= 1\times\Bigl(\cos\frac{3\pi}{4}+i\,\sin\frac{3\pi}{4}\Bigr)\end{align*}

   and it follows that

   \displaystyle \begin{align*}&\Bigl(\frac{1}{\sqrt2} -\frac{i}{\sqrt2}\Bigr) \Big/ \Bigl(-\frac{1}{\sqrt2} +\frac{i}{\sqrt2}\Bigr) = \smash{\frac{\cos\dfrac{7\pi}{4}+i\,\sin\dfrac{7\pi}{4}\vphantom{\Biggl(}}{\cos\dfrac{3\pi}{4}+i\,\sin\dfrac{3\pi}{4}\vphantom{\Biggl)}}}\\[16pt] &\qquad\quad{}= \cos\Bigl(\frac{7\pi}{4}-\frac{3\pi}{4}\Bigl)+i\,\sin\Bigl(\frac{7\pi}{4}-\frac{3\pi}{4}\Bigr)= \cos\pi+i\,\sin\pi=-1\,\mbox{.}\end{align*}




2. \displaystyle (-2-2i)(1+i)
   
   The factors in the expression are written in polar form

   \displaystyle \begin{align*}-2-2i&=\sqrt8\Bigl(\cos\frac{5\pi}{4}+i\,\sin\frac{5\pi}{4}\Bigr)\,\mbox{,}\\[4pt] 1+i&=\sqrt2\Bigl(\cos\frac{\pi}{4}+i\,\sin\frac{\pi}{4}\Bigr)\,\mbox{.}\end{align*}

   Multiplication in polar form gives

   \displaystyle \begin{align*}(-2-2i)(1+i)&=\sqrt8 \times \sqrt2\,\Bigl(\cos\Bigl(\frac{5\pi}{4}+\frac{\pi}{4}\Bigr)+i\,\sin\Bigl(\frac{5\pi}{4}+\frac{\pi}{4}\Bigr)\Bigr)\\[4pt] &=4\Bigl(\cos\frac{3\pi}{2}+i\,\sin\frac{3\pi}{2} \Bigr)=-4i\,\mbox{.}\end{align*}

Example 7

1. Simplify \displaystyle iz and \displaystyle \frac{z}{i} if \displaystyle \ z=2\Bigl(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\Bigr). Answer in polar form.
   
   Since \displaystyle \ i=1\times \left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\ it follows that

   \displaystyle \begin{align*} iz &= 2\Bigl(\cos\Bigl(\frac{\pi}{6}+\frac{\pi}{2}\Bigr)+i\,\sin\Bigl(\frac{\pi}{6}+\frac{\pi}{2}\Bigr)\,\Bigr)= 2\Bigl(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\Bigr)\,\mbox{,}\\[4pt] \frac{z}{i} &= 2\Bigl(\cos\Bigl(\frac{\pi}{6}-\frac{\pi}{2}\Bigr)+i\,\sin\Bigl(\frac{\pi}{6}-\frac{\pi}{2}\Bigr)\,\Bigr) = 2\Bigl(\cos\frac{-\pi}{3}+i\,\sin\frac{-\pi}{3}\Bigr)\,\mbox{.}\end{align*}




2. Simplify \displaystyle iz and \displaystyle \frac{z}{i} if \displaystyle \ z=3\left(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\right)\,. Answer in polar form.
   
   Rewriting \displaystyle i in polar form gives

   \displaystyle \begin{align*} iz &= 3\Bigl(\cos\Bigl(\frac{7\pi}{4}+\frac{\pi}{2}\Bigr)+i\,\sin\Bigl(\frac{7\pi}{4}+\frac{\pi}{2}\Bigr)\,\Bigr) = 3\Bigl(\cos\frac{9\pi}{4}+i\sin\frac{9\pi}{4}\Bigr)\\[4pt] &= 3\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\,\mbox{,}\\[6pt] \frac{z}{i} &= 2\Bigl(\cos\Bigl(\frac{7\pi}{4}-\frac{\pi}{2}\Bigr)+i\,\sin\Bigl(\frac{7\pi}{4}-\frac{\pi}{2}\Bigr)\,\Bigr)= 2\Bigl(\cos\frac{5\pi}{4}+i\,\sin\frac{5\pi}{4}\Bigr)\,\mbox{.}\end{align*}

We see that multiplying by i leads to an anticlockwise rotation \displaystyle \pi/2, while division with i results in a clockwise rotation \displaystyle \pi/2.