Complex roots

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#1 h May 15, 2019, 1:12 PM • 1 Y

Y by Adventure10

2.(a)Count the number of all the complex roots $\omega$ of the equation $z^{2019}-1=0$ which follows $|\omega+1|\geq \sqrt{2+\sqrt{2}}$ (b) Solve for real $x$
$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$

This post has been edited 3 times. Last edited by Sarbajit10598, May 18, 2019, 7:14 AM

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Figaro

#2 h May 15, 2019, 1:45 PM • 2 Y

Y by Adventure10, Mango247

Hint for (b): factorize the left side.

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#3 h May 15, 2019, 2:08 PM • 3 Y

Y by Farbe_Bears, Adventure10, Mango247

Let $u=(\frac{3}{2})^x$ then it reduces to $\frac{u^3+1}{u+u^2}=\frac{7}{6}$ Solving for $u=-1,\frac{2}{3},\frac{3}{2}$ .The last two values give $x=+1,-1$ which are only real solution.

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#4 h May 15, 2019, 2:13 PM • 2 Y

Y by Adventure10, Mango247

(b) The equation it is: $6\cdot {{2}^{3x}}+6\cdot {{3}^{3x}}=7\cdot {{2}^{2x}}\cdot {{3}^{x}}+7\cdot {{2}^{x}}\cdot {{3}^{2x}}\Leftrightarrow 6\left( {{\left( \frac{2}{3} \right)}^{3x}}+1 \right)=7{{\left( \frac{2}{3} \right)}^{x}}\left( {{\left( \frac{2}{3} \right)}^{x}}+1 \right)$ , now denote ${{\left( \frac{2}{3} \right)}^{x}}=t$ , so we get $t=-1$ or $6(t^2-t+1)=7t$ , and the rest is easy.

This post has been edited 1 time. Last edited by oldepsilon, May 15, 2019, 2:14 PM

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#5 h May 15, 2019, 2:48 PM • 2 Y

Y by Sarbajit10598, Adventure10

Notation : $\text{cis}(x) = \cos(x) + i {\sin(x)}$ .

Also , we note that $1 + \text{cis}(2x) = 1 + \cos(2x) + i {\sin(2x)} = 2{\cos}^2(x) + 2i{\cos(x)}{\sin(x)} = 2{\cos(x)}{\text{cis}(x)}$ .

And so , $|1 + \text{cis}(2x)| = 2|\cos(x)|$ .

Now $z^{2019} = 1 \implies z = \text{cis}\left(\frac{2k{\pi}}{2019}\right)$ where $k \in \{0 , 1 , 2 , \cdots , 2018\}$ .

Let $y = \frac{k{\pi}}{2019}$ so that we have $0 \le y < {\pi}$

We have $|z + 1| = |1 + \text{cis}(2y)| = 2|\cos(y)|$

So , we have ,

$2|\cos(y)| \ge \sqrt{2+\sqrt{2}}$

$\implies |\cos(y)| \ge \frac{\sqrt{2+\sqrt{2}}}{2}$

$\implies {\cos}^2(y) \ge \frac{2+\sqrt{2}}{4}$

$\implies \frac{1+\cos(2y)}{2} \ge \frac{2+\sqrt{2}}{4}$

$\implies \cos(2y) \ge \frac{1}{\sqrt{2}}$

$\implies 2y \in \left[0 , \frac{\pi}{4}\right] \cup \left[\frac{7{\pi}}{4} , 2{\pi}\right)$

$\implies y \in \left[0 , \frac{\pi}{8}\right] \cup \left[\frac{7{\pi}}{8} , {\pi}\right)$

$\implies \frac{k{\pi}}{2019} \in \left[0 , \frac{\pi}{8}\right] \cup \left[\frac{7{\pi}}{8} , {\pi}\right)$

$\implies k \in \left[0 , \frac{2019}{8}\right] \cup \left[\frac{14133}{8} , 2019\right)$ .

So $k \in [0 , 252.3] \cup [1766.6 , 2018]$

So , the answer is $505$ .

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#6 h Jun 17, 2020, 2:05 PM

Y by

Sarbajit10598 wrote:

2.(a)Count the number of all the complex roots $\omega$ of the equation $z^{2019}-1=0$ which follows $|\omega+1|\geq \sqrt{2+\sqrt{2}}$ (b) Solve for real $x$
$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$

a nicer solution to (a) I think.
We get $(\omega+1)(\overline{\omega}+1) \ge 2+\sqrt{2}$
or $\frac{\omega+\overline{\omega}}{2} \ge \frac{1}{\sqrt{2}}$
or $\Re(\omega) \ge \frac{1}{\sqrt{2}}$
which when you see on the unit circle translates to finding all 2019th roots of unity $z$ such that $|\text{arg}(z)| \le \frac{\pi}{4}$ , which in turn is just the number of integer solutions $k$ to $\frac{2 \pi k}{2019} \le \frac{\pi }{4}$ which is just $2 \left \lfloor{\frac{2019}{8}}\right \rfloor+1 = \boxed{505}$

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#8 h Oct 21, 2020, 10:24 AM

Y by

MathBoy23 wrote:

Sarbajit10598 wrote:

2.(a)Count the number of all the complex roots $\omega$ of the equation $z^{2019}-1=0$ which follows $|\omega+1|\geq \sqrt{2+\sqrt{2}}$ (b) Solve for real $x$
$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$

a nicer solution to (a) I think.
We get $(\omega+1)(\overline{\omega}+1) \ge 2+\sqrt{2}$
or $\frac{\omega+\overline{\omega}}{2} \ge \frac{1}{\sqrt{2}}$
or $\Re(\omega) \ge \frac{1}{\sqrt{2}}$
which when you see on the unit circle translates to finding all 2019th roots of unity $z$ such that $|\text{arg}(z)| \le \frac{\pi}{4}$ , which in turn is just the number of integer solutions $k$ to $\frac{2 \pi k}{2019} \le \frac{\pi }{4}$ which is just $2 \left \lfloor{\frac{2019}{8}}\right \rfloor+1 = \boxed{505}$

Sorry for reviving!

I wanted to know why have you added 1 in the end?

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#9 h Friday at 5:57 PM

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@above probably to count $k =0$

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