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Number of roots of the equation $z^{10}-z^5-992=0$ where real parts are negative is

(a) 3

(b)4

(c)5

(d)6

What I've tried so far

Let $z=x+iy$

Now, putting the value of $z$ in the equation, we get:

But, this is leading me to a very big equation. I am not quite sure how to solve this equation.

Any help or hint would be much awaited and appreciated.

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  • $\begingroup$ was that supposed to be $z^{10}$ or $0 \cdot z^1$? $\endgroup$ – Siddharth Bhat Jun 10 '16 at 7:02
  • $\begingroup$ I am sorry! Can you edit it to the first one? $\endgroup$ – Abhishekstudent Jun 10 '16 at 7:03
  • $\begingroup$ sure thing, Will do :) FYI, the way to write it that way is to write z^{10} $\endgroup$ – Siddharth Bhat Jun 10 '16 at 7:03
  • $\begingroup$ Thanks for your help, sir! $\endgroup$ – Abhishekstudent Jun 10 '16 at 7:05
  • $\begingroup$ It looks much much better, sir! Thanks for your precious time! $\endgroup$ – Abhishekstudent Jun 10 '16 at 7:06
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Replace $z^5 \to t$. Hence, this gives us $t^2 - t - 992 = 0$. Solving for the roots (use the quadratic equation) we get that the roots are

$t = 32, -31$

Hence, $z^5 = 32, -31$

For

$$ z^5 = 32 \\ z^5 = e^{i 2\pi n}2^5, n \in \mathbb{N} \\ z = 2 \cdot e^{\frac{i 2 \pi n}{5}}, n \in \mathbb{N} $$

Hence, we need to think about the roots laid out in the complex plane with an angle of $\frac{360^\circ}{5} = 72^\circ$ between them. The roots with negtive real part will lie between $90^\circ \leq \theta \leq 270^\circ$ - that is, the roots at $144^\circ$ and $216^\circ$.

Similary, analyze $$ z^5 = -31 $$

By symmetry, since the exponent is the same as the previous case ($5$), we will have the same anglular distance - $72^\circ$. However, in this case, it is flipped about the $y$-axis (the imaginary axis) since we have a $-1$ factor. Hence, this time, we will have $3$ negative roots (since the last time we have $3$ positive roots).

So, the total is $3 + 2 = 5$ roots with a negative real part

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Hint

$$z^{10}-z^5-992=0\implies Z^2-Z-992=0\implies(Z+31)(Z-32)=0$$ using $Z=z^5$.

I am sure that you can take it from here.

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The equation is quadratic in $z^5$ with one positive and one negative root. $$ z^5=a\implies z=a^{1\over5}e^{2\pi i{k\over5}}, 0\le k<5\\ \Re z=a^{1\over5}\cos(2\pi k/5) $$ If $a>0,$ we have to make the cosine part negative which leaves us with $k=2,3$.

If $a<0,$ we have to make the cosine part positive which leaves us with $k=0,1,4$.

So there are $5$ roots with negative real part.

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