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