Question

If $11{z^{10}} + 10i{z^9} + 10iz - 11 = 0$, then what will be the possible value of $\left| z \right|$.

Answer 1

The given function is a polynomial of degree $10$ and so it is analytic and the equation has $10$ roots. One can solve this equation by rearranging the equation by keeping the higher order on one side. Then we can prove by showing the contradiction $\left| z \right| < 1$or $\left| z \right| > 1$. Finally we will get the required result.

Complete step-by-step solution:
Consider the given equation $11{z^{10}} + 10i{z^9} + 10iz - 11 = 0$.
Rearranging the equation by keeping the higher order on one side and the remaining on the other side.
${z^9}\left( {11z + 10i} \right) = 11 - 10iz$
Taking the term $11z + 10i$ to other side,
$\Rightarrow {z^9} = \dfrac{{11 - 10iz}}{{11z + 10i}} - - - - \left( 1 \right)$
Let $z$ be a complex number that satisfies the equation $11{z^{10}} + 10i{z^9} + 10iz - 11 = 0$, $z = x + iy$, we can find the roots by considering two cases.
Case 1:
For $\left| z \right| < 1$,
$\Rightarrow \left| {x + iy} \right| = \sqrt {\left( {x + iy} \right)\left( {x - iy} \right)}$
$\Rightarrow \sqrt {{x^2} + {y^2}} < 1$
Taking modulus on both sides of the equation (1), we get
$\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10iz}}{{11z + 10i}}{\text{ }}} \right|$
Substitute the value of z,
$\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10i\left( {x + iy} \right)}}{{11\left( {x + iy} \right) + 10i}}} \right|$
Expanding further we get,
$\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10ix + 10y}}{{11x + 11iy + 10i}}} \right| - - - - \left( 2 \right)$
We know that $\left| {a + ib} \right| = \sqrt {\left( {a + ib} \right)\left( {a - ib} \right)} = \sqrt {{a^2} + {b^2}}$
Consider the numerator $\left| {11 - 10ix + 10y} \right|$,
Let $a = 11 + 10y,b = - 10x$
$\left| {11 - 10ix + 10y} \right| = \sqrt {{{\left( {10y + 11} \right)}^2} + {{\left( { - 10x} \right)}^2}}$
$\Rightarrow \sqrt {{{\left( {10y + 11} \right)}^2} + 100{x^2}} - - - - \left( 3 \right)$
Consider the denominator $\left| {11x + 11iy + 10i} \right|$,
Let $a = 11x,b = 11y + 10$
$\left| {11x + 11iy + 10i} \right| = \sqrt {{{\left( {11x} \right)}^2} + {{\left( {11y + 10} \right)}^2}}$
$\Rightarrow \sqrt {{{\left( {11y + 10} \right)}^2} + 121{x^2}} - - - - \left( 4 \right)$
Substituting equation (3) and (4) in (2),
$\left| {{z^9}} \right| = \dfrac{{\sqrt {{{\left( {10y + 11} \right)}^2} + 100{x^2}} {\text{ }}}}{{\sqrt {{{\left( {11y + 10} \right)}^2} + 121{x^2}} }}$
Expanding sum of squares on both numerator and denominator, we get
$\Rightarrow \left| {{z^9}} \right| = \dfrac{{\sqrt {100{y^2} + 220y + 121 + 100{x^2}} {\text{ }}}}{{\sqrt {121{y^2} + 220y + 100 + 121{x^2}} }}$
Hence,
$\Rightarrow \left| {{z^9}} \right| = \dfrac{{\sqrt {100\left( {{x^2} + {y^2}} \right) + 220y + 121} {\text{ }}}}{{\sqrt {121\left( {{x^2} + {y^2}} \right) + 220y + 100} }}$
$\because {x^2} + {y^2} < 1$
Rearranging the terms we get,
$\therefore 100\left( {{x^2} + {y^2}} \right) + 220y + 121 > 121\left( {{x^2} + {y^2}} \right) + 220y + 100$
Hence,
$\Rightarrow \left| {{z^9}} \right| > 1$
$\Rightarrow \left| z \right| > 1$
This is a contradiction.
Case 2:
For $\left| z \right| > 1$,
$\Rightarrow \left| {x + iy} \right| = \sqrt {\left( {x + iy} \right)\left( {x - iy} \right)}$
$\Rightarrow \sqrt {{x^2} + {y^2}} > 1$
Taking modulus on both sides of the equation (1), we get
$\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10iz}}{{11z + 10i}}{\text{ }}} \right|$
Substitute the value of z,
$\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10i\left( {x + iy} \right)}}{{11\left( {x + iy} \right) + 10i}}} \right|$
Expanding further we get,
$\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10ix + 10y}}{{11x + 11iy + 10i}}} \right| - - - - \left( 5 \right)$
We know that $\left| {a + ib} \right| = \sqrt {\left( {a + ib} \right)\left( {a - ib} \right)} = \sqrt {{a^2} + {b^2}}$
Consider the numerator $\left| {11 - 10ix + 10y} \right|$,
Let $a = 11 + 10y,b = - 10x$
$\left| {11 - 10ix + 10y} \right| = \sqrt {{{\left( {10y + 11} \right)}^2} + {{\left( { - 10x} \right)}^2}}$
$\Rightarrow \sqrt {{{\left( {10y + 11} \right)}^2} + 100{x^2}} - - - - \left( 6 \right)$
Consider the denominator $\left| {11x + 11iy + 10i} \right|$,
Let $a = 11x,b = 11y + 10$
$\left| {11x + 11iy + 10i} \right| = \sqrt {{{\left( {11x} \right)}^2} + {{\left( {11y + 10} \right)}^2}}$
$\Rightarrow \sqrt {{{\left( {11y + 10} \right)}^2} + 121{x^2}} - - - - \left( 7 \right)$
Substituting equation (6) and (7) in (5),
$\left| {{z^9}} \right| = \dfrac{{\sqrt {{{\left( {10y + 11} \right)}^2} + 100{x^2}} {\text{ }}}}{{\sqrt {{{\left( {11y + 10} \right)}^2} + 121{x^2}} }}$
Expanding sum of squares on both numerator and denominator, we get
$\Rightarrow \left| {{z^9}} \right| = \dfrac{{\sqrt {100{y^2} + 220y + 121 + 100{x^2}} {\text{ }}}}{{\sqrt {121{y^2} + 220y + 100 + 121{x^2}} }}$
Simplifying we get,
$\Rightarrow \left| {{z^9}} \right| = \dfrac{{\sqrt {100\left( {{x^2} + {y^2}} \right) + 220y + 121} {\text{ }}}}{{\sqrt {121\left( {{x^2} + {y^2}} \right) + 220y + 100} }}$
$\because {x^2} + {y^2} > 1$
Rearranging the terms we get,
$\therefore 100\left( {{x^2} + {y^2}} \right) + 220y + 121 > 121\left( {{x^2} + {y^2}} \right) + 220y + 100$
Hence,
$\Rightarrow \left| {{z^9}} \right| < 1$
$\Rightarrow \left| z \right| < 1$
This is a contradiction.
$\therefore$ $\left| z \right| = 1$, which means all the roots lie on the circle.

Note: In logic and mathematics, proof by contradiction is a form of proof that establishes the truth or the validity of a proposition, by showing that assuming the proposition to be false leads to a contradiction. Proof by contradiction is also known as indirect proof, proof by assuming the opposite.