[출제의도] 켤레복소수의 성질을 이용하여 곱셈공식의 변형 이해하기
z = a + b i z = a + bi z = a + bi a , b a , b a , b z ‾ = a − b i \displaystyle \overline{z} = a - bi z = a − bi z 2 = ( a + b i ) 2 = 3 + 4 i z ^{2} = ( a + bi ) ^{2} = 3 + 4 i z 2 = ( a + bi ) 2 = 3 + 4 i a 2 − b 2 + 2 a b i = 3 + 4 i a ^{2} - b ^{2} + 2 abi = 3 + 4 i a 2 − b 2 + 2 abi = 3 + 4 i a 2 − b 2 = 3 , a b = 2 a ^{2} - b ^{2} = 3 , ab = 2 a 2 − b 2 = 3 , ab = 2 ∴ z z ‾ \displaystyle \therefore z \overline{z} ∴ z z = ( a + b i ) ( a − b i ) = ( a + bi ) ( a - bi ) = ( a + bi ) ( a − bi ) = a 2 + b 2 = a ^{2} + b ^{2} = a 2 + b 2 ( a 2 + b 2 ) 2 \displaystyle \sqrt{( a ^{2} + b ^{2} ) ^{2}} ( a 2 + b 2 ) 2 = ( a 2 − b 2 ) 2 + 4 a 2 b 2 \displaystyle = \sqrt{( a ^{2} - b ^{2} ) ^{2} + 4 a ^{2} b ^{2}} = ( a 2 − b 2 ) 2 + 4 a 2 b 2 = 3 2 + 4 × 2 2 = 5 \displaystyle = \sqrt{3 ^{2} + 4 \times 2 ^{2}} = 5 = 3 2 + 4 × 2 2 = 5 ( z z ‾ ) 2 = z 2 ( z ‾ ) 2 = z 2 z 2 ‾ = 25 \displaystyle ( z \overline{z} ) ^{2} = z ^{2} {( \overline{z} ) ^{2} = z ^{2} \overline{z ^{2}} = 25} ( z z ) 2 = z 2 ( z ) 2 = z 2 z 2 = 25 ∴ z z ‾ = 5 ( ∵ z z ‾ ≥ 0 ) \displaystyle \therefore z \overline{z} = 5 ( \because z \overline{z} \geq 0 ) ∴ z z = 5 ( ∵ z z ≥ 0 )