IBDP Maths AHL 1.13 Modulus-argument in polar form AA HL Paper 2- Exam Style Questions- New Syllabus

(a)

Using Euler’s formula \( e^{i\phi} = \cos\phi + i\sin\phi \), we know that \( \sin\phi = \operatorname{Im}(e^{i\phi}) \).
Therefore:
\( f(t) = \operatorname{Im}(e^{i(2t+1)}) \)
\( g(t) = \operatorname{Im}(e^{i(2t+3)}) \)
\( h(t) = \operatorname{Im}(e^{i(2t+1)} + e^{i(2t+3)}) \)
Factoring out \( e^{2it} \):
\( h(t) = \operatorname{Im}(e^{2it}e^{i} + e^{2it}e^{3i}) = \operatorname{Im}\left( e^{2it}(e^{i} + e^{3i}) \right) \).
Shown.

(b)

To write \( z = e^{i} + e^{3i} \) in form \( re^{i\theta} \), we can factor out the “average” exponent:
\( z = e^{2i}(e^{-i} + e^{i}) \)
Using the identity \( \cos\phi = \frac{e^{i\phi} + e^{-i\phi}}{2} \), we have \( e^{i} + e^{-i} = 2\cos 1 \).
Thus, \( z = (2\cos 1)e^{2i} \).

Since \( \cos 1 \approx 0.5403 > 0 \), the modulus \( r = 2\cos 1 \approx 1.08 \).
The argument \( \theta = 2 \).
Answer: \( \boxed{r = 2\cos 1, \theta = 2} \)

(c)

Substitute the result from (b) into the expression from (a):
\( h(t) = \operatorname{Im}\left( e^{2it} \cdot (2\cos 1)e^{2i} \right) \)
\( h(t) = \operatorname{Im}\left( 2\cos 1 \cdot e^{i(2t+2)} \right) \)
Using the definition of the imaginary part:
\( h(t) = 2\cos 1 \sin(2t + 2) \).
Here \( p = 2\cos 1 \approx 1.08 \) and \( q = 2 \).
Answer: \( \boxed{h(t) = 2\cos 1 \sin(2t+2)} \)