Solve: ∣∣(1+i)(3+i)(2+i)∣∣=

∣∣(1+i)(3+i)(2+i)∣∣=∣∣(1+i)(3+i)(2+i)3−i3−i∣∣=32−(i)2(i2+3i+2)(3−i)=10(i2+3i+2)(3−i)=10(3i+1)(3−i)=109i+3−3i2−i=108i+6=54i+3

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Solve: |(1+i){(2+i)}{(3+i)}|=

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Solve: $∣ ∣ (1 + i) \frac{( 2 + i )}{( 3 + i )} ∣ ∣ =$

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∣ ∣ (1 + i) \frac{( 2 + i )}{( 3 + i )} ∣ ∣ =

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∣ ∣ (1 + i) \frac{( 2 + i )}{( 3 + i )} ∣ ∣ = ∣ ∣ (1 + i) \frac{( 2 + i )}{( 3 + i )} \frac{3 - i}{3 - i} ∣ ∣ = \frac{( i ^{2} + 3 i + 2 ) ( 3 - i )}{3 ^{2} - ( i ) ^{2}} = \frac{( i ^{2} + 3 i + 2 ) ( 3 - i )}{10} = \frac{( 3 i + 1 ) ( 3 - i )}{10} = \frac{9 i + 3 - 3 i ^{2} - i}{10} = \frac{8 i + 6}{10} = \frac{4 i + 3}{5}

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