Polar form arithmetic 
Complex numbers in polar form, especially when expressed in exponential form
r e i θ , {r\mathrm{e}^{\mathrm{i}\theta},} r e i θ ,   lends itself especially well to multiplication, division  
and conjugation  .
( r 1 e i θ 1 ) ⋅ ( r 2 e i θ 2 ) = ( r 1 r 2 ) e i ( θ 1 + θ 2 ) ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ arg  ( z 1 z 2 ) = arg  ( z 1 ) + arg  ( z 2 ) \begin{aligned}
    (r_1\mathrm{e}^{\mathrm{i}\theta_1}) \cdot (r_2\mathrm{e}^{\mathrm{i}\theta_2}) &= (r_1r_2)\mathrm{e}^{\mathrm{i}(\theta_1 + \theta_2)} \\
    \left| z_1 z_2 \right| &= \left| z_1 \right| \left| z_2 \right| \\
    \arg(z_1 z_2) &= \arg(z_1) + \arg(z_2)
  \end{aligned} ( r 1  e i θ 1  ) ⋅ ( r 2  e i θ 2  ) ∣ z 1  z 2  ∣ arg  ( z 1  z 2  )  = ( r 1  r 2  ) e i ( θ 1  + θ 2  ) = ∣ z 1  ∣ ∣ z 2  ∣ = arg  ( z 1  ) + arg  ( z 2  )   
r 1 e i θ 1 r 2 e i θ 2 = r 1 r 2 e i ( θ 1 − θ 2 ) ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ arg  ( z 1 z 2 ) = arg  ( z 1 ) − arg  ( z 2 ) {\begin{aligned}
    \frac{r_1\mathrm{e}^{\mathrm{i}\theta_1}}{r_2\mathrm{e}^{\mathrm{i}\theta_2}} &= \frac{r_1}{r_2}\mathrm{e}^{\mathrm{i}(\theta_1 - \theta_2)} \\
    \left| \frac{z_1}{z_2} \right| &= \frac{\left| z_1 \right|}{\left| z_2 \right|} \\
    \arg\left(\frac{z_1}{z_2}\right) &= \arg\left(z_1\right) - \arg\left(z_2\right)
  \end{aligned}} r 2  e i θ 2  r 1  e i θ 1   ∣ ∣  z 2  z 1   ∣ ∣  arg  ( z 2  z 1   )  = r 2  r 1   e i ( θ 1  − θ 2  ) = ∣ z 2  ∣ ∣ z 1  ∣  = arg  ( z 1  ) − arg  ( z 2  )   
( r e i θ ) ∗ = r e − i θ ∣ z ∗ ∣ = ∣ z ∣ arg  ( z ∗ ) = − arg  ( z ) \begin{aligned}
    (r\mathrm{e}^{\mathrm{i}\theta})^* &= r\mathrm{e}^{-\mathrm{i}\theta} \\
    \left| z^* \right| &= \left| z \right| \\
    \arg(z^*) &= -\arg(z)
  \end{aligned} ( r e i θ ) ∗ ∣ z ∗ ∣ arg  ( z ∗ )  = r e − i θ = ∣ z ∣ = − arg  ( z )   
Principal argument 
The problem with angles are that there are more than one way to express the same angle (by going
multiple rounds). For example, the angles 1 4 π \frac{1}{4}\pi 4 1  π   represents the same angle as 9 4 π \frac{9}{4}\pi 4 9  π  
and − 3 4 π -\frac{3}{4}\pi − 4 3  π  .
We resolve this issue by defining the principal argument of a complex number to be between − π -\pi − π   and π \pi π  .
Principal argument: − π < θ ≤ π . {-\pi < \theta \leq \pi.} − π < θ ≤ π . 
 
If any of our arguments are outside of this range, we can add or subtract 2 k π 2k\pi 2 k π   from it to bring it back.
Examples 
Evaluate 2 e 2 3 π i × 3 e 5 6 π i {2 \mathrm{e}^{ \frac{2}{3} \pi \mathrm{i} } \times 3 \mathrm{e}^{ \frac{5}{6} \pi \mathrm{i} }} 2 e 3 2  π i × 3 e 6 5  π i  
	
	( 2 e 2 3 π i ) ⋅ ( 3 e 5 6 π i ) z = ( 2 ⋅ 3 ) e i ( 2 3 π + 5 6 π ) z = 6 e 3 2 π i 
			\begin{aligned}
				& \left(2 \mathrm{e}^{ \frac{2}{3} \pi \mathrm{i} }\right) \cdot \left(3 \mathrm{e}^{ \frac{5}{6} \pi \mathrm{i} }\right) \\
					\phantom{z} &= \left(2\cdot3\right) \mathrm{e}^{\mathrm{i}\left( \frac{2}{3} \pi   + \frac{5}{6} \pi  \right)} \\
					\phantom{z} &= 6\mathrm{e}^{\frac{3}{2} \pi\mathrm{i}}
			\end{aligned}
		 z z  ( 2 e 3 2  π i ) ⋅ ( 3 e 6 5  π i ) = ( 2 ⋅ 3 ) e i ( 3 2  π + 6 5  π ) = 6 e 2 3  π i