Using complex analysis to convert $b\cos \theta +a \sin \theta$ to a single trigonometric function

Question

Using product $(a+bi)(\cos \theta+i \sin \theta) $ show that $$b\cos \theta +a \sin \theta=\sqrt{a^2 + b^2}\sin(\theta+\arctan(b/a))$$ and using this result show by induction that $$ \frac{\operatorname{d}^n}{\operatorname{d}^nt}(e^{at}\sin (bt))=(a^2 +b^2)^{\frac{n}{2}}e^{at} \sin(bt+n \arctan(b/a))$$ Any hints on how to tackle this are welcome.

Answer 1

$$ r(\cos\theta + i \sin \theta) = r\mathrm{e}^{i\theta} $$ and we can write any complex number in expoential form with the correct $(r,\theta)$ $$ a+ib = r\mathrm{e}^{i\phi} = r(\cos \phi + i \sin \phi) $$

so $$ r\cos\phi = a,\\ r \sin \phi = b $$ or $$ \tan \phi = \dfrac{b}{a}\implies \phi = \arctan \left(\dfrac{b}{a}\right) $$

now $$ r^2\cos^2\phi + r^2\sin^2\phi = a^2+b^2 = r^2\implies r = \sqrt{a^2+b^2} $$

so we can write $$ a + ib = \sqrt{a^2+b^2}\mathrm{e}^{i\arctan \left(\dfrac{b}{a}\right)} = $$

the second brackets is readily in the form to make use of euler as $$ \mathrm{e}^{i\theta} $$ so we end up at $$ (a+ib)(\cos \theta + i \sin \theta) = \sqrt{a^2+b^2}\mathrm{e}^{i\tan^{-1}\left(b/a\right)}\mathrm{e}^{i\theta} $$ or $$ \sqrt{a^2+b^2}\mathrm{e}^{i\left(\tan^{-1}\left(b/a\right) + \theta\right)}\tag{*} $$

expanding the original product we get $$ a\cos \theta -b\sin \theta + ib\cos \theta + ia\sin\theta $$ where the imaginary component is $$ b\cos \theta + a \sin\theta $$ so now we know to obtain the the indenity we have to take the imaginary component of Eq.(*) which is $$ \sqrt{a^2+b^2}\sin\left(\tan^{-1}\left(b/a\right) + \theta\right) $$