ㄱ. ω = a + b i \omega = a + b i ω = a + bi 라 하면 z = ω ‾ = a − b i \displaystyle z = \overline{\omega} = a - b i z = ω = a − bi
∣ z + ω ∣ = ∣ 2 a ∣ = 2 ∣ a ∣ | z + \omega | = | 2 a | = 2 | a | ∣ z + ω ∣ = ∣2 a ∣ = 2∣ a ∣ 이고
2 ∣ z ∣ = 2 a 2 + b 2 \displaystyle 2 | z | = 2 \sqrt{a ^{2} + b ^{2}} 2∣ z ∣ = 2 a 2 + b 2
2 ∣ a ∣ ≤ 2 a 2 + b 2 \displaystyle 2 | a | \leq 2 \sqrt{a ^{2} + b ^{2}} 2∣ a ∣ ≤ 2 a 2 + b 2 이므로 ∣ a ∣ ≤ a 2 + b 2 \displaystyle | a | \leq \sqrt{a ^{2} + b ^{2}} ∣ a ∣ ≤ a 2 + b 2
ㄴ. ω = a + b i \omega = a + b i ω = a + bi 라 하면 z = i ω = a i − b z = i \omega = a i - b z = iω = ai − b
∣ z − ω ∣ = ∣ ( − b − a ) + ( a − b ) i ∣ | z - \omega | = | ( - b - a ) + ( a - b ) i | ∣ z − ω ∣ = ∣ ( − b − a ) + ( a − b ) i ∣
= ( − b − a ) 2 + ( a − b ) 2 \displaystyle = \sqrt{( - b - a ) ^{2} + ( a - b ) ^{2}} = ( − b − a ) 2 + ( a − b ) 2
= 2 a 2 + b 2 = 2 a 2 + b 2 = 2 ∣ z ∣ \displaystyle = 2 \sqrt{a ^{2} + b ^{2}} = \sqrt{2} \sqrt{a ^{2} + b ^{2}} = \sqrt{2} | z | = 2 a 2 + b 2 = 2 a 2 + b 2 = 2 ∣ z ∣
ㄷ. ω = a + b i \omega = a + b i ω = a + bi 라 하면 z = − ω ‾ = − a + b i \displaystyle z = - \overline{\omega} = - a + b i z = − ω = − a + bi
∣ z + ω ∣ = ∣ 2 b i ∣ = 2 b 2 = 2 ∣ b ∣ \displaystyle | z + \omega | = | 2 b i | = 2 \sqrt{b ^{2}} = 2 | b | ∣ z + ω ∣ = ∣2 bi ∣ = 2 b 2 = 2∣ b ∣
∣ z ∣ 2 = ( a 2 + b 2 ) 2 = a 2 + b 2 \displaystyle | z | ^{2} = \left( \sqrt{a ^{2} + b ^{2}} \right) ^{2} = a ^{2} + b ^{2} ∣ z ∣ 2 = ( a 2 + b 2 ) 2 = a 2 + b 2
만약, a = 1 4 , b = 1 3 \displaystyle a = \frac{1}{4} , b = \frac{1}{3} a = 4 1 , b = 3 1 이면 2 ∣ b ∣ > a 2 + b 2 2 | b | > a ^{2} + b ^{2} 2∣ b ∣ > a 2 + b 2 이므로 성립하지 않는다.