复数的一些性质

1. $|c{|}^2=c\bar{c}$

$证明$：
$$\begin{array}{rl}|c{|}^2& =a^2+b^2\\ & =(a+ib)(a-ib)\end{array}$$

2. $\overline{c_1{c}_2}={\overline{c}}_1{\overline{c}}_1$

$证明$：
$$\begin{array}{rl}\overline{c_1{c}_2}& =\overline{(a_1+i{b}_1)(a_2+i{b}_2)}\\ & \\ & =\overline{a_1{a}_2-b_1{b}_2+i(a_1{b}_1+a_2{b}_2)}\\ & \\ & =a_1{a}_2-b_1{b}_2-i(a_1{b}_1+a_2{b}_2)\\ & \\ & =(a_1-i{b}_1)(a_2-i{b}_2)\\ & \\ & ={\overline{c}}_1{\overline{c}}_1\end{array}$$

3. ∣ c 1 − c 2 ∣ 2 = ∣ c 1 ∣ 2 + ∣ c 2 ∣ 2 − 2 ℜ { c 1 c ‾ 2 } = ∣ c 1 ∣ 2 + ∣ c 2 ∣ 2 − 2 ℜ { c 2 c ‾ 1 } |c_1 - c_2|^2 = |c_1|^2 + |c_2|^2 -2\Re\{c_1 \overline{c}_2\} = |c_1|^2 + |c_2|^2 -2\Re\{c_2 \overline{c}_1\} ∣c1​−c2​∣2=∣c1​∣2+∣c2​∣2−2ℜ{ c1​c2​}=∣c1​∣2+∣c2​∣2−2ℜ{ c2​c1​}

$证明$：
∣ c 1 − c 2 ∣ 2 = ∣ a 1 − a 2 + i ( b 1 − b 2 ) ∣ 2 = ( a 1 − a 2 ) 2 + ( b 1 − b 2 ) 2 = a 1 2 + b 1 2 + a 2 2 + b 2 2 − 2 ( a 1 a 2 + b 1 b 2 ) = ∣ c 1 ∣ 2 + ∣ c 2 ∣ 2 − 2 ℜ { c 1 c ˉ 2 } = ∣ c 1 ∣ 2 + ∣ c 2 ∣ 2 − 2 ℜ { c 2 c ˉ 1 } \begin{aligned} |c_1 - c_2|^2 &= |a_1-a_2 +i(b_1-b_2)|^2 \\ &= (a_1-a_2)^2 + (b_1-b_2)^2 \\ &= a_1^2+b_1^2 + a_2^2 + b_2^2 -2(a_1a_2+b_1b_2) \\ &= |c_1|^2 + |c_2|^2 -2\Re\{c_1 \bar{c}_2\} \\ &= |c_1|^2 + |c_2|^2 -2\Re\{c_2 \bar{c}_1\} \end{aligned} ∣c1​−c2​∣2​=∣a1​−a2​+i(b1​−b2​)∣2=(a1​−a2​)2+(b1​−b2​)2=a12​+b12​+a22​+b22​−2(a1​a2​+b1​b2​)=∣c1​∣2+∣c2​∣2−2ℜ{ c1​cˉ2​}=∣c1​∣2+∣c2​∣2−2ℜ{ c2​cˉ1​}​