Question

The distance of the roots of the equation $|\sin {\theta }_1|z^3+|\sin {\theta }_2|z^2+|\sin {\theta }_3|z+|\sin {\theta }_4|=3$, from $z=0$ are

• A. Greater than $\frac{2}{3}$
• B. Less than $\frac{2}{3}$
• C. Greater than $|\sin {\theta }_1|+|\sin {\theta }_2|+|\sin {\theta }_3|+|\sin {\theta }_4|$
• D. Less than $|\sin {\theta }_1|+|\sin {\theta }_2|+|\sin {\theta }_3|+|\sin {\theta }_4|$

Answer 1

The correct option is A Greater than $\frac{2}{3}$

Let $z=z_1+i{z}_2$ and $w=w_1+i{w}_2$ be two complex numbers. Distance between these two complex numbers on the complex plane can be found using the Pythagoras theorem as follows-

Distance = $\sqrt{{(z_1-w_1)}^2+{(z_2-w_2)}^2}$

=$\left|\right(z_1-w_1)+i(z_2-w_2\left)\right|$ {Since $|x+iy|=\sqrt{x^2+y^2}$}

=$\left|\right(z_1+i{z}_2)-(w_1+i{w}_2\left)\right|$

=$|z-w|$

This shows that the distance between any two complex numbers is the modulus of their difference.

The given equation is $\left|\sin {\theta }_1\right|z^3+\left|\sin {\theta }_2\right|z^2+\left|\sin {\theta }_3\right|z+\left|\sin {\theta }_4\right|=3$

Hence to find the distance between the roots of this equation and $z=0$ we have to find the modulus of the difference of $z$ and $0$, i.e.- we have to find the value of $|z-0|$=$\left|z\right|$

We know that for any value of ${\theta }_n$ , $\left|\sin {\theta }_n\right|\le 1$

Now, taking modulus on both sides of the given equation we get-

$|\left|\sin {\theta }_1\right|z^3+\left|\sin {\theta }_2\right|z^2+\left|\sin {\theta }_3\right|z+\left|\sin {\theta }_4\right||=\left|3\right|$

$\Rightarrow 3=|\left|\sin {\theta }_1\right|z^3+\left|\sin {\theta }_2\right|z^2+\left|\sin {\theta }_3\right|z+\left|\sin {\theta }_4\right||$

$\Rightarrow 3\le |z^3+z^2+z+1|$ {Using $\left|\sin {\theta }_n\right|\le 1$}

$\Rightarrow 3\le \left|z^3\right|+\left|z^2\right|+\left|z\right|+\left|1\right|$ {Using Triangle Inequality}

$\Rightarrow 3<1+\left|z\right|+\left|z^2\right|+\left|z^3\right|+..........\infty$

$\Rightarrow 3<1+\left|z\right|+{\left|z\right|}^2+{\left|z\right|}^3+..........\infty$

$\Rightarrow 3<\frac{1}{1-\left|z\right|}$ {Using the sum of infinite GP series and the fact that $\left|z\right|<1$}

$\Rightarrow 1-\left|z\right|<\frac{1}{3}$

$\Rightarrow 1-\frac{1}{3}<\left|z\right|$

$\Rightarrow \left|z\right|>\frac{2}{3}$

This shows the distance between the roots of the given equation and $z=0$ is greater than $\frac{2}{3}$.

Hence, the correct option is Option A.