Let S be the set of all (,),

Question

Let $S$ be the set of all $(\alpha,\beta)$, $\pi<\alpha,\beta<2\pi$, for which the complex number $\dfrac{1-i\sin\alpha}{1+2i\sin\alpha}$ is purely imaginary and $\dfrac{1+i\cos\beta}{1-2i\cos\beta}$ is purely real. Let $Z_{\alpha\beta}=\sin2\alpha+i\cos2\beta,\ (\alpha,\beta)\in S$. Then $\displaystyle\sum_{(\alpha,\beta)\in S}\left(iZ_{\alpha\beta}+\dfrac1{iZ_{\alpha\beta}}\right)$ is equal to:

Accepted Answer

Purely imaginary condition: adding the expression to its conjugate and setting to zero gives $1-2\sin^2\alpha=0 \implies \sin\alpha=\pm\dfrac1{\sqrt2}$. In $(\pi,2\pi)$: $\alpha=\dfrac{5\pi}4,\dfrac{7\pi}4$.

Purely real condition: similarly gives $\cos\beta=0$, so $\beta=\dfrac{3\pi}2$ in $(\pi,2\pi)$.

$$S=\left\{\left(\frac{5\pi}4,\frac{3\pi}2\right),\left(\frac{7\pi}4,\frac{3\pi}2\right)\right\}.$$

For these, $Z_{\alpha\beta}=1-i$ and $Z_{\alpha\beta}=-1-i$ respectively.

$$\sum\left(iZ+\frac1{iZ}\right) = i(-2i)+\frac1i\cdot\frac1{1+i}+\frac1{-1+i} = 2+\frac1i\left(\frac{2i}{-2}\right)=1.$$

Solutions · 1