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# Number of roots of the equation ${{z}^{10}}-{{z}^{5}}-992=0$ where real parts are negative isA. $3$B. $4$C. $5$D. $6$

Last updated date: 13th Sep 2024
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Hint: Convert ${{z}^{10}}-{{z}^{5}}-992=0$ to $a{{x}^{2}}+bx+c=0$, for that let us assume ${{z}^{5}}=t$. After that, find the roots of the equation and then find the number of roots.

Complete step-by-step solution -

Now here we want to find the number of roots of the equation ${{z}^{10}}-{{z}^{5}}-992=0$ .
We are aware that we can solve the equation in the form $a{{x}^{2}}+bx+c=0$ and can find the number of roots.
So we have to convert ${{z}^{10}}-{{z}^{5}}-992=0$ to $a{{x}^{2}}+bx+c=0$ , for that let us assume ${{z}^{5}}=t$,
So the equation becomes,
${{t}^{2}}-t-992=0$
So solving above quadratic equation we get,

\begin{align} & {{t}^{2}}-32t+31t-992=0 \\ & t(t-32)+31(t-32)=0 \\ & (t-32)(t+31)=0 \\ \end{align}
So $(t-32)=0$or $(t+31)=0$
So $t=32,-31$
Now we know ${{z}^{5}}=t$ , So re-substituting we get,
So ${{z}^{5}}=32$ or ${{z}^{5}}=-31$
So we get, $z={{\left( 32 \right)}^{\dfrac{1}{5}}}$ and $z={{\left( -31 \right)}^{\dfrac{1}{5}}}$ .
So Now $z={{\left( 32 \right)}^{\dfrac{1}{5}}}={{({{2}^{5}})}^{\dfrac{1}{5}}}$
So $z=2{{e}^{i\left( \dfrac{0+2n\pi }{5} \right)}}$ ……….. (The argument of $32$ is $0$ )
And the value of n will lie from $0$ to $4$ .
So for $n=0$ we get,
$z=2{{e}^{0}}=2$……….. (1)
For $n=1$ we get,
$z=2{{e}^{i\left( \dfrac{2\pi }{5} \right)}}$…………. (2)
For $n=2$ we get,

$z=2{{e}^{i\left( \dfrac{4\pi }{5} \right)}}$….. (3)
For $n=3$ we get,
$z=2{{e}^{i\left( \dfrac{6\pi }{5} \right)}}$…………… (4)
For $n=4$ we get,
$z=2{{e}^{i\left( \dfrac{8\pi }{5} \right)}}$……….. (5)
Similarly $z={{\left( -31 \right)}^{\dfrac{1}{5}}}$
So $z={{(31)}^{\dfrac{1}{5}}}{{e}^{i\left( \dfrac{\pi +2n\pi }{5} \right)}}$ ……….. (The argument of $-31$ is $\pi$ )
So let assume ${{\left( 31 \right)}^{\dfrac{1}{5}}}=r$
So at $n=0$We get,
$z=r{{e}^{i\left( \dfrac{\pi +2(0)\pi }{5} \right)}}=r{{e}^{i\left( \dfrac{\pi }{5} \right)}}$…………. (6)
For $n=1$ we get,
$z=r{{e}^{i\left( \dfrac{\pi +2(1)\pi }{5} \right)}}=r{{e}^{i\left( \dfrac{3\pi }{5} \right)}}$………… (7)
For $n=2$ we get,
$z=r{{e}^{i\left( \dfrac{\pi +2(2)\pi }{5} \right)}}=r{{e}^{i\left( \pi \right)}}$……………. (8)
For $n=3$ we get,
$z=r{{e}^{i\left( \dfrac{\pi +2(3)\pi }{5} \right)}}=r{{e}^{i\left( \dfrac{7\pi }{5} \right)}}$…………….. (9)
For $n=4$ we get,
$z=r{{e}^{i\left( \dfrac{\pi +2(4)\pi }{5} \right)}}=r{{e}^{i\left( \dfrac{9\pi }{5} \right)}}$……………….. (10)
According to the question we are asked that the real part should be negative.
So the condition for the real part to be negative is, the argument should lie between $\dfrac{\pi }{2}$ to $\dfrac{3\pi }{2}$ .
So the arguments which lie between $\dfrac{\pi }{2}$ to $\dfrac{3\pi }{2}$ are (3), (4), (7), (8) and (9).
So the number of roots is $5$.

Note: So basically first you have to read the question and understand what is asked. Here the number of roots are $5$ but our total roots are $10$ .So if you do not read the question properly then there are chances that you may write it as $10$. You should be clear that the real part to be negative means the argument should lie between $\dfrac{\pi }{2}$ to $\dfrac{3\pi }{2}$ . This concept should be clear.