Formulas and Results of Complex Numbers

$z = (a,b) = a + ib,{\text{ }}i = (0,1)$

$i = \sqrt { - 1} ,{\text{ }}{i^2} =- 1,{\text{}}{i^3} =- i,{\text{ }}{i^4} = 1,{\text{ }}{i^5} = i,{\text{ }} \ldots$

If $n$ is a positive integer, then ${(i)^{4n}} = 1,{\text{ }}{(i)^{4n + 1}} = i,{\text{ }}{(i)^{4n + 2}} =- 1,{\text{}}{(i)^{4n + 3}} =- i$

If $a + ib = 0$ then $a = b = 0$, and conversely

If $a + ib = c + id$ then $a = c$ and $b = d$

$(a,b) + (c,d) = (a + c,{\text{ }}b + d)$

$(a,b)(c,d) = (ac - bd,{\text{ }}ad + bc)$

${z_1} + {z_2} = {z_2} + {z_1}{\text{ ;}}\forall {z_1},{z_2} \in \mathbb{C}$

${z_1} \cdot {z_2} = {z_2} \cdot {z_1}{\text{ ;}}\forall {z_1},{z_2} \in \mathbb{C}$

${z_1} + ({z_2} + {z_3}) = ({z_1} + {z_2}{\text{) + }}{{\text{z}}_{\text{3}}}{\text{ ;}}\forall {z_1},{z_2},{z_3} \in \mathbb{C}$

${z_1} \cdot ({z_2} \cdot {z_3}) = ({z_1} \cdot {z_2}{\text{)}} \cdot {{\text{z}}_{\text{3}}}{\text{ ;}}\forall {z_1},{z_2},{z_3} \in \mathbb{C}$

$(0,0)$ is the additive identity.

$(1,0)$ is the multiplicative identity.

If $z = a + ib$ the multiplicative inverse of $z$ is ${z^{ - 1}} = \frac{a}{{{a^2} + {b^2}}} - i\frac{b}{{{a^2} + {b^2}}}$

The additive inverse of $z$ is $- z =- a - ib$

If $z = a + ib$, then $\overline z = a - ib$

$\overline {{z_1} + {z_2}}= \overline {{z_1}} + \overline {{z_2}} {\text{ ;}}\forall {z_1},{z_2} \in \mathbb{C}$

$\overline {{z_1} - {z_2}}= \overline {{z_1}} - \overline {{z_2}} {\text{ ; }}\forall {z_1},{z_2} \in \mathbb{C}$

$\overline {{z_1} \cdot {z_2}}= \overline {{z_1}}\cdot \overline {{z_2}} {\text{ ; }}\forall {z_1},{z_2} \in \mathbb{C}$

$\overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)}= \frac{{\overline {{z_1}} }}{{\overline {{z_2}} }}{\text{ ; }}\forall {z_1},{z_2} \in \mathbb{C}$

If $\overline z = z$, then $z$ is a real number.

$\overline {\left( {\overline z } \right)}= z$

If $z = a + ib$, $a = \operatorname{Re} (z),{\text{ }}b = \operatorname{Im} (z)$

$z{\text{ }}\overline z= {(\operatorname{Re} z)^2} + {(\operatorname{Im} z)^2}$

If $z = a + ib$, the $\left| z \right| = \sqrt {{a^2} + {b^2}}$

$\left| z \right| \geqslant 0$

$\left| z \right| = \left| { - z} \right| = \left| {\overline z } \right|$

${\left| z \right|^2} = z{\text{ }}\overline z$

$\left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right|$

$\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}{\text{, }}{z_2} \ne 0$

$\left| {{z_1}} \right| - \left| {{z_2}} \right| \leqslant \left| {{z_1} + {z_2}} \right| \leqslant \left| {{z_1}} \right| + \left| {{z_2}} \right|$

$\left| {{z_1} - {z_2}} \right| \geqslant \left| {{z_1}} \right| - \left| {{z_2}} \right|$

$\left| {\operatorname{Re} z} \right| \leqslant \left| z \right|$, $\left| {\operatorname{Im} z} \right| \leqslant \left| z \right|$

$\left| {{z_1} - {z_2}} \right| = \left| {{z_2} - {z_1}} \right|$

$\left| {\left| {{z_1}} \right| - \left| {{z_2}} \right|} \right| \leqslant \left| {{z_1} - {z_2}} \right|$

$z = r(Cos\theta + iSin\theta )$ is polar form of $z$, where $r = \left| z \right|{\text{ ;}}\theta = Ta{n^{ - 1}}\left( {\frac{b}{a}} \right) = \arg (z)$

If ${z_1} = {r_1}(Cos{\theta _1} + iSin{\theta _1})$ and ${z_2} = {r_2}(Cos{\theta _2} + iSin{\theta _2})$, then