Question

Find the value of a belongs to real numbers for which ai is a solution of z^4-2z^3+7z^2-4z+10=0 also find all the roots of this equation

Answer 1

Answer on Question #67466 – Math – Complex Analysis

Question

Find the value of $a \in \mathbb{R}$ for which $ai$ is a solution of

$$z^4 - 2z^3 + 7z^2 - 4z + 10 = 0.$$

Also find all the roots of this equation.

Solution

Let us substitute $ai$ into the equation $z^4 - 2z^3 + 7z^2 - 4z + 10 = 0$ given that

i^2 = -1; \quad i^3 = -i; \quad i^4 = 1 \quad \text{(see https://en.wikipedia.org/wiki/Complex_number)}.

We get

$$a^4 + 2a^3i - 7a^2 - 4ai + 10 = 0 \Leftrightarrow (a^4 - 7a^2 + 10) + (2a^3 - 4a)i = 0.$$

From the definition of equality of two complex numbers (see http://www.math-only-math.com/equality-of-complex-numbers.html) we conclude that

$$\left\{ \begin{array}{c} a^4 - 7a^2 + 10 = 0 \\ 2a^3 - 4a = 0 \end{array} \right.$$

Let us solve the second equation of the system:

$$2a(a^2 - 2) = 0 \Leftrightarrow \left\{ \begin{array}{c} a = 0 \\ a = \sqrt{2} \\ a = -\sqrt{2} \end{array} \right.$$

But $a = 0$ does not satisfy the first equation.

Let us check $a = \sqrt{2} \colon 4 - 7 \cdot 2 + 10 = 0 \Rightarrow a = \sqrt{2}$ is a solution of the obtained system.

Let us check $a = -\sqrt{2} \colon 4 - 7 \cdot 2 + 10 = 0 \Rightarrow a = -\sqrt{2}$ is a solution of the obtained system.

So we have two roots of the original equation: $\left\{

\begin{array}{l}

z_1 = i\sqrt{2} \\

z_2 = -i\sqrt{2}

\end{array}

\right.$ (corresponding to the values

$$\left\{ \begin{array}{c} a_1 = \sqrt{2} \\ a_2 = -\sqrt{2} \end{array} \right.$$

From the polynomial remainder theorem

(see https://en.wikipedia.org/wiki/Polynomial_remainder_theorem) it follows that the polynomial $z^4 - 2z^3 + 7z^2 - 4z + 10$ is divisible by polynomial $(z - i\sqrt{2})(z + i\sqrt{2}) = z^2 + 2$. Using long polynomial division (see https://en.wikipedia.org/wiki/Polynomial_long_division) we obtain

$$\frac {z ^ {4} - 2 z ^ {3} + 7 z ^ {2} - 4 z + 10}{z ^ {2} + 2} = z ^ {2} - 2 z + 5.$$

To solve the equation $z^2 - 2z + 5 = 0$ we apply the quadratic formula (see https://en.wikipedia.org/wiki/Quadratic_formula) and obtain that

$$\left\{ \begin{array}{l} z _ {3} = 1 + 2 i \\ z _ {4} = 1 - 2 i \end{array} \right.$$

So all the roots of the original equation are

$$\left\{ \begin{array}{l} z _ {1} = i \sqrt {2} \\ z _ {2} = - i \sqrt {2} \\ z _ {3} = 1 + 2 i \\ z _ {4} = 1 - 2 i \end{array} \right.$$

Answer: $a = \pm \sqrt{2}; z = \pm i\sqrt{2}, z = 1 \pm 2i.$

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