Question

Let $a,b,c$ be three complex numbers satisfying the equation $|z|=1.$ If $a+b\cos \alpha +c\sin \alpha =0$, where $\alpha \in (0,\frac{\pi }{2})$, then which of the following is incorrect?

• A. $b^2+c^2=0$
• B. $b^2-c^2=0$
• C. $a=-b{e}^{i\alpha }$
• D. $a=-b{e}^{-i\alpha }$

Answer 1

The correct option is B $b^2-c^2=0$

Given : $a+b\cos \alpha +c\sin \alpha =0$
$\Rightarrow -a=b\cos \alpha +c\sin \alpha \Rightarrow |-a{|}^2=(b\cos \alpha +c\sin \alpha )(\overline{b}\cos \alpha +\overline{c}\sin \alpha )\Rightarrow |a{|}^2=|b{|}^2{\cos }^2\alpha +|c{|}^2{\sin }^2\alpha +(b\overline{c}+c\overline{b})\sin \alpha \cos \alpha \Rightarrow 1=1+(\frac{b}{c}+\frac{c}{b})\sin \alpha \cos \alpha (\because |a|=|b|=|c|=1)\Rightarrow (\frac{b}{c}+\frac{c}{b})\sin \alpha \cos \alpha =0\Rightarrow b^2+c^2=0(\alpha \in (0,\frac{\pi }{2}))\Rightarrow c=\pm bi\Rightarrow a=-(b\cos \alpha \pm bi\sin \alpha )\therefore a=-b{e}^{\pm i\alpha }$