QCE Vault / Specialist Maths Specialist Maths — Question 13 QCAA 2021, Paper 1 · 6 marks
Browse all questions Original practice exam Report an issue Q13 · 2021 · Technology-free Simple familiar 6 marks
QUESTION 13 (6 marks) Use z = a + b i z=a+bi z = a + bi and w = c + d i w=c+di w = c + d i , where a , b , c , d ∈ R a,b,c,d\in\mathbb R a , b , c , d ∈ R , to prove
∣ z − w ∣ 2 = ∣ z ∣ 2 + ∣ w ∣ 2 − 2 Re ( z w ‾ ) . |z-w|^2=|z|^2+|w|^2-2\operatorname{Re}(z\overline w). ∣ z − w ∣ 2 = ∣ z ∣ 2 + ∣ w ∣ 2 − 2 Re ( z w ) . WORKED SOLUTION
QCAA guide · typeset solution 6 marks Worked solution
z − w = ( a − c ) + ( b − d ) i z-w=(a-c)+(b-d)i z − w = ( a − c ) + ( b − d ) i , so ∣ z − w ∣ 2 = ( a − c ) 2 + ( b − d ) 2 = a 2 + b 2 + c 2 + d 2 − 2 ( a c + b d ) |z-w|^2=(a-c)^2+(b-d)^2=a^2+b^2+c^2+d^2-2(ac+bd) ∣ z − w ∣ 2 = ( a − c ) 2 + ( b − d ) 2 = a 2 + b 2 + c 2 + d 2 − 2 ( a c + b d ) . Also ∣ z ∣ 2 + ∣ w ∣ 2 = a 2 + b 2 + c 2 + d 2 |z|^2+|w|^2=a^2+b^2+c^2+d^2 ∣ z ∣ 2 + ∣ w ∣ 2 = a 2 + b 2 + c 2 + d 2 , while z w ‾ = ( a + b i ) ( c − d i ) = ( a c + b d ) + ( b c − a d ) i z\overline w=(a+bi)(c-di)=(ac+bd)+(bc-ad)i z w = ( a + bi ) ( c − d i ) = ( a c + b d ) + ( b c − a d ) i , so Re ( z w ‾ ) = a c + b d \operatorname{Re}(z\overline w)=ac+bd Re ( z w ) = a c + b d . Substitution gives the required identity. correctly expresses z − w z-w z − w in terms of a , b , c , d a,b,c,d a , b , c , d in Cartesian form [1 mark] expresses ∣ z − w ∣ 2 |z-w|^2 ∣ z − w ∣ 2 in terms of a , b , c , d a,b,c,d a , b , c , d in expanded form [1 mark] correctly expresses ∣ z ∣ 2 + ∣ w ∣ 2 |z|^2+|w|^2 ∣ z ∣ 2 + ∣ w ∣ 2 in terms of a , b , c , d a,b,c,d a , b , c , d in expanded form [1 mark] correctly expresses Re ( z w ‾ ) \operatorname{Re}(z\overline w) Re ( z w ) in terms of a , b , c , d a,b,c,d a , b , c , d [1 mark] shows logical organisation, communicating key steps
[1 mark] One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.
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