Q259 · Practice questionTechnology-freeComplex familiar5 marks
QUESTION 259 (5 marks)
Let , with . You may use and the modulus product identity.
a)[3 marks]
Prove without assuming the quotient identity.
b)[2 marks]
Hence prove for , and state why the restriction matters.
WORKED SOLUTION
5 marksPractice marking scheme
ANSWER
(a) . (b) ; zero has no multiplicative inverse.
Worked solution
(a) Conjugating a quotient gives . Therefore
Since , taking nonnegative square roots gives the result, including .
(b) Apply part (a) with numerator 1 and denominator . Both and denote the same inverse. For , these expressions and are undefined.
Equivalent justified methods accepted; respect any requested proof method.
Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.
Part a: Use the conjugate of a quotient.
Part a: Obtain .
Part a: Take nonnegative roots and state why division is valid.
Part b: Apply the established quotient result to .
Part b: Explain the exclusion of zero.
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