复数的一些性质

2023-12-27 02:54| 来源: 网络整理| 查看: 265

∣ c ∣ 2 = c c ˉ |c|^2 = c\bar{c} ∣c∣2=ccˉ

证明 \colorbox{red}{证明} 证明： ∣ c ∣ 2 = a 2 + b 2 = ( a + i b ) ( a − i b ) \begin{aligned} |c|^2 &= a^2 + b^2 \\ &= (a+ib)(a-ib) \end{aligned} ∣c∣2=a2+b2=(a+ib)(a−ib)

c 1 c 2 ‾ = c ‾ 1 c ‾ 1 \overline{c_1 c_2} = \overline{c}_1\overline{c}_1 c1c2=c1c1

证明 \colorbox{red}{证明} 证明： c 1 c 2 ‾ = ( a 1 + i b 1 ) ( a 2 + i b 2 ) ‾ = 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 ) = c ‾ 1 c ‾ 1 \begin{aligned} \overline{c_1 c_2} &= \overline{(a_1 + i b_1) (a_2 + ib_2)} \\\\ &= \overline{a_1a_2 -b_1 b_2 +i(a_1b_1 + a_2 b_2)} \\\\ &= a_1a_2 -b_1 b_2 -i(a_1b_1 + a_2 b_2) \\\\ &= (a_1 - i b_1) (a_2 - ib_2) \\\\ &= \overline{c}_1\overline{c}_1 \end{aligned} c1c2=(a1+ib1)(a2+ib2)=a1a2−b1b2+i(a1b1+a2b2)=a1a2−b1b2−i(a1b1+a2b2)=(a1−ib1)(a2−ib2)=c1c1

∣ 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{c1c2}=∣c1∣2+∣c2∣2−2ℜ{c2c1}

证明 \colorbox{red}{证明} 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(a1a2+b1b2)=∣c1∣2+∣c2∣2−2ℜ{c1cˉ2}=∣c1∣2+∣c2∣2−2ℜ{c2cˉ1}

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