Special complex numbers
In cartesian form, a complex number can be expressed as z = x + y i . z=x+y\mathrm{i}. z = x + y i . y y y real x x x purely imaginary
Cartesian form
Condition
Cartesian form,x + y i {x+yi} x + y i Real π΅π’
y = 0 {y=0} y = 0 Real and positive π΅
y = 0 , x > 0 y=0, \allowbreak x > 0 y = 0 , x > 0 Real and negative π’
y = 0 , x < 0 y=0, \allowbreak x < 0 y = 0 , x < 0 Purely imaginary π΄
x = 0 {x = 0} x = 0
Example
Given z = 3 + b i , {z=3+b\mathrm{i},} z = 3 + b i , b {b} b b {b} b z 2 z β {\displaystyle \frac{z^2}{z^*}} z β z 2 β
Step 1:
Evaluate z 2 z β {\displaystyle \frac{z^2}{z^*}} z β z 2 β
z 2 z β = ( 3 + b i ) 2 3 β b i = ( 3 + b i ) 2 3 β b i β
3 + b i 3 + b i = 3 ( 9 β b 2 ) β 6 b 2 + ( 18 b + ( 9 β b 2 ) b ) i 9 + b 2 = 27 β 9 b 2 + b ( 27 β b 2 ) i 9 + b 2 \begin{aligned}
\frac{z^2}{z^*} &= \frac{(3+b\mathrm{i})^2}{3-b\mathrm{i}} \\
&= \frac{(3+b\mathrm{i})^2}{3-b\mathrm{i}} \cdot \frac{3+b\mathrm{i}}{3+b\mathrm{i}} \\
&= \frac{3(9-b^2)-6b^2+(18b+(9-b^2)b)\mathrm{i}}{9+b^2} \\
&= \frac{27-9b^2+b(27-b^2)\mathrm{i}}{9+b^2}
\end{aligned} z β z 2 β β = 3 β b i ( 3 + b i ) 2 β = 3 β b i ( 3 + b i ) 2 β β
3 + b i 3 + b i β = 9 + b 2 3 ( 9 β b 2 ) β 6 b 2 + ( 18 b + ( 9 β b 2 ) b ) i β = 9 + b 2 27 β 9 b 2 + b ( 27 β b 2 ) i β β Polar form
In polar form, multiple answers/arguments could be possible due to βextra roundsβ
bringing us out of the principal range. To account for this, we have an
additional k Ο k\pi k Ο 2 k Ο 2k\pi 2 k Ο k k k
Condition
Polar form,r e i ΞΈ {re^{i\theta}} r e i ΞΈ Real π΅π’
ΞΈ = k Ο {{\theta = k \pi}} ΞΈ = kΟ Real and positive π΅
ΞΈ = 2 k Ο {{\theta = 2k \pi}} ΞΈ = 2 kΟ Real and negative π’
ΞΈ = Ο + 2 k Ο {{\theta = \pi + 2k \pi}} ΞΈ = Ο + 2 kΟ Purely imaginary π΄
ΞΈ = Ο 2 + k Ο {\theta = \frac{\pi}{2} + k \pi} ΞΈ = 2 Ο β + kΟ k β Z {k \in \mathbb{Z}} k β Z
Example
Find the three smallest positive integers
n {n} n ( e β Ο 12 i ) n {\displaystyle \left(\mathrm{e}^{-\frac{\pi}{12}\mathrm{i}}\right)^n} ( e β 12 Ο β i ) n
Step 1:
Purely imaginary β arg β‘ ( z ) = Ο 2 + k Ο {\Rightarrow \arg(z) = \frac{\pi}{2} + k\pi} β arg ( z ) = 2 Ο β + kΟ
β n Ο 12 = Ο 2 + k Ο whereΒ k β Z n = β 6 β 12 k \begin{gather*}
-\frac{n\pi}{12} = \frac{\pi}{2} + k\pi \\
\textrm{where } k \in \mathbb{Z} \\
n = -6 - 12k
\end{gather*} β 12 nΟ β = 2 Ο β + kΟ whereΒ k β Z n = β 6 β 12 k β