Complex Numbers / Series
Derivatives of real and imaginary parts of e^{xe^{iα}}
MJAT_TS8_P2
Grade 12
Question:
Consider $p=x\sin\alpha+\frac{x^2}{2!}\sin2\alpha+\frac{x^3}{3!}\sin3\alpha+\cdots$ and $q=1+x\cos\alpha+\frac{x^2}{2!}\cos2\alpha+\cdots$ Then:
A) $\left(\frac{dp}{dx}\right)^2+\left(\frac{dq}{dx}\right)^2=p^2+q^2$
B) $\left(\frac{dp}{dx}\right)^2+\left(\frac{dq}{dx}\right)^2=q^2$
C) If $\int_0^1(p^2+q^2)dx=f(\alpha)$ then $f(0)=\dfrac{e^2-1}{2}$
D) If $\int_0^1(p^2+q^2)dx=f(\alpha)$ then $f(0)=\dfrac{e^2+1}{2}$
Step-by-Step Solution
Key Concept: $q+ip=e^{xe^{i\alpha}}$. Differentiating: $\frac{dq}{dx}+i\frac{dp}{dx}=e^{i\alpha}\cdot e^{xe^{i\alpha}}=e^{i\alpha}(q+ip)$. So $|dq/dx+i\,dp/dx|^2=|q+ip|^2$, giving $(dq/dx)^2+(dp/dx)^2=p^2+q^2$ (A ✓).
A ✓, C ✓. Answer: A, C.
Correct Answer: AC