Mathematics - Complex Number Question with Solution | TestHub

Given,

$x^2-\sqrt{6}x+3=0$ has roots $\alpha \& \beta$

Now, solving $x^2-\sqrt{6}x+3=0$ we get,

$x=\frac{\sqrt{6}\pm i\sqrt{6}}{2}=\sqrt{3}\left(\frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i\right)$

Now, taking $\alpha =\sqrt{3}\left(e^{i\frac{\pi }{4}}\right), \beta =\sqrt{3}\left(e^{-i\frac{\pi }{4}}\right)$

Now, solving $\frac{{\alpha }^{99}}{\beta }+{\alpha }^{98}={\alpha }^{98}\left(\frac{\alpha }{\beta }+1\right)$

$=\frac{{\alpha }^{98}(\alpha +\beta )}{\beta }$

$=\frac{3^{49}{\left(e^{i\frac{\pi }{4}}\right)}^{98}\left(\sqrt{6}\right)}{\sqrt{3}\left(e^{-i\frac{\pi }{4}}\right)}$

$=3^{49}\left(e^{i99\frac{\pi }{4}}\right)\times \sqrt{2}$

$=3^{49}\left(\cos \frac{99\pi }{4}+i\sin \frac{99\pi }{4}\right)\times \sqrt{2}$

$=3^{49}\left(\cos \left(25\pi -\frac{\pi }{4}\right)+i\sin \left(25\pi -\frac{\pi }{4}\right)\right)\times \sqrt{2}$

$=3^{49}(-1+i)$

$=3^n(a+ib)$

So, on comparing we get,

$n=49, a=-1, b=1$

$\therefore n+a+b=49-1+1=49$