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TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Section: Mathematics

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Question : 9 of 160

Marks: +1, -0

If the roots of the equation

(z-4)^3

8i

a-2i,b+i

c+i

\sqrt{abc}

=

$13\sqrt{3}$
$4\sqrt{13}$
$2\sqrt{13}$
$5\sqrt{3}$

Solution: 👈: Video Solution

z-4=(8i)^{\frac{1}{3}}

z=4+2(i)^{\frac{1}{3}}

i=e^{i\pi/2}=e^{i\left(\frac{\pi}{2}+2n\pi\right)}

i^{\frac{1}{3}}=e^{i\left(\frac{\pi}{6}+\frac{2n\pi}{3}\right)}, n=0,1,2

\begin{bmatrix} e^{i\frac{\pi}{6}} = \cos\frac{\pi}{6}+i\sin\frac{\pi}{6} & = \frac{\sqrt{3}}{2}+i\frac{1}{2} \\ e^{i\left(\frac{\pi}{6}+\frac{2\pi}{3}\right)} & = e^{i\frac{5\pi}{6}} = \cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6} = \frac{-\sqrt{3}}{2}+i\frac{1}{2} \\ e^{i\left(\frac{\pi}{6}+\frac{4\pi}{3}\right)} & = e^{i\frac{3\pi}{2}} \\ & = \cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}=0-i \end{bmatrix}

For,

n=0, e^{i\pi/6}=\frac{\sqrt{3}}{2}+i\frac{1}{2}

n=1, e^{i5\pi/6}=-\frac{\sqrt{3}}{2}+i\cdot\frac{1}{2}

n=2, e^{i\frac{3\pi}{2}}=0-i

\therefore z=4+2\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)=(4+\sqrt{3})+i=b+i

z=4+2\left(\frac{-\sqrt{3}}{2}+\frac{i}{2}\right)=(4-\sqrt{3})+i=c+i

\Rightarrow z=4+2(0-i)=4-2i=a-2i

\therefore a=4, b=4+\sqrt{3} \text{ and } c=4-\sqrt{3}

\therefore 4(4+\sqrt{3})(4-\sqrt{3}) = \sqrt{4(16-3)} = 2\sqrt{13}

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