• # question_answer If the complex number is ${{\left( 1+ri \right)}^{3}}=\lambda \left( 1+i \right),$ when  $i=\sqrt{-1},$ for some real $\lambda$, the value of $r$ can be A) $\cos \frac{\pi }{5}$ B) $\operatorname{cosec}\frac{3\pi }{2}$ C) $\cos \frac{\pi }{12}$ D) $\operatorname{cosec}\frac{\pi }{12}$

 $\therefore$${{(1+ri)}^{3}}=\lambda (1+i)$$\Rightarrow$$1-{{r}^{3}}i+3ri-3{{r}^{2}}=\lambda +i\lambda$ On comparing real and imaginary parts. We get $1-3{{r}^{2}}=\lambda$and $-{{r}^{3}}+3r=\lambda$ Then $-{{r}^{3}}-3r=1-3{{r}^{2}}$ $\Rightarrow$${{r}^{3}}-3{{r}^{2}}-3r+1=0$ $\Rightarrow$$({{r}^{3}}+1)-3r(r+1)=0$ $\Rightarrow$$(r+1)({{r}^{2}}-r+1-3r)=0$ $\Rightarrow$$(r+1)({{r}^{2}}-4r+1)=0\therefore r=-1,2\pm \sqrt{3}$$\Rightarrow$$r=\operatorname{cosec}\frac{3\pi }{2}$