Proof by mathematical induction — Question 15

QCAA 2025, Paper 2 · 6 marks

Q15 · 2025 · Technology-activeComplex familiar6 marks

QUESTION 15 (6 marks)

De Moivre’s theorem can be expressed as

\bigl(r(\cos\theta+i\sin\theta)\bigr)^n=r^n\bigl(\cos(n\theta)+i\sin(n\theta)\bigr)\qquad\forall n\in\mathbb Z^+.

Prove De Moivre’s theorem using mathematical induction.

WORKED SOLUTION

6 marks

ANSWER

De Moivre’s theorem is proved for all

n\in\mathbb Z^+

Worked solution

Initial statement: for

n=1

r(\cos\theta+i\sin\theta)=r^1(\cos\theta+i\sin\theta),

so the result is true.

Assume it is true for

n=k

\bigl(r(\cos\theta+i\sin\theta)\bigr)^k=r^k(\cos(k\theta)+i\sin(k\theta)).

For

n=k+1

\bigl(r(\cos\theta+i\sin\theta)\bigr)^{k+1} =r^k(\cos(k\theta)+i\sin(k\theta))\,r(\cos\theta+i\sin\theta).

Expanding in Cartesian form gives

r^{k+1}\bigl(\cos(k\theta)\cos\theta-\sin(k\theta)\sin\theta+i(\sin(k\theta)\cos\theta+\cos(k\theta)\sin\theta)\bigr).

Using angle-sum identities,

=r^{k+1}\bigl(\cos((k+1)\theta)+i\sin((k+1)\theta)\bigr).

Thus the statement is true for

k+1

, so by mathematical induction it is true for

n=1,2,\ldots

correctly proves the initial statement

[1 mark]

correctly states a suitable assumption

[1 mark]

uses the assumption statement

[1 mark]

expresses the result in Cartesian form

[1 mark]

uses angle sum and difference identities

[1 mark]

completes the proof and states a suitable conclusion

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

How many marks did you earn?

Compare your working with the guide above.