Chapter 8.3, Problem 100E

(a)

To determine

To solve: given equation.

Answer to Problem 100E

The roots of $z^2+\left(1+i\right)z+i=0$ are $-iand-1$

Explanation of Solution

Given:

  $z^2+\left(1+i\right)z+i=0$

Calculation:

Consider $z^2+\left(1+i\right)z+i=0$

Find the roots of $z^2+\left(1+i\right)z+i=0$using the quadratic formula.

Use De Moivre’s Theorem to rewrite the roots. Recall that for a polynomial $a{x}^2+bx+c$ , the quadratic formula gives the roots:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In this case, $a=1,\text{ b}=1+i,and\text{ c}=i$ .

Thus, the quadratic formula gives:

  $\begin{array}{l}x=\frac{-\left(1+i\right)\pm \sqrt{{\left(1+i\right)}^2-4i}}{2}\\ =\frac{-\left(1+i\right)\pm \sqrt{1+2i+i^2-4i}}{2}\\ =\frac{-\left(1+i\right)\pm \sqrt{1-2i+i^2}}{2}\\ =\frac{-\left(1+i\right)\pm \sqrt{{\left(1-i\right)}^2}}{2}\\ =\frac{-1-i\pm \left(1-i\right)}{2}\end{array}$

  $\begin{array}{l}Forx=\frac{-1-i+\left(1-i\right)}{2}\\ x=\frac{-1-i+\left(1-i\right)}{2}\\ =\frac{-2i}{2}\\ =-i\end{array}$

  $\begin{array}{l}Forx=\frac{-1-i-\left(1-i\right)}{2}\\ x=\frac{-1-i-\left(1-i\right)}{2}\\ =\frac{-2}{2}\\ =-1\end{array}$

Hence, the roots of $z^2+\left(1+i\right)z+i=0$ are $-iand-1$

De Moivre’s Theorem is not necessary.

Conclusion:

Hence, the roots of $z^2+\left(1+i\right)z+i=0$ are $-iand-1$

(b)

To determine

To solve: given equation.

Answer to Problem 100E

The roots of $z^2-iz+1=0$ are $\frac{i+i\sqrt{5}}{2}and\frac{i-i\sqrt{5}}{2}$ .

Explanation of Solution

Given:

  $z^2-iz+1=0$

Calculation:

Consider $z^2-iz+1=0$

Find the roots of $z^2-iz+1=0$using the quadratic formula. Recall that for a polynomial $a{x}^2+bx+c$ , the quadratic formula gives the roots:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In this case, $a=1,\text{ b}=-i,and\text{ c}=1$ . Thus, the quadratic formula gives:

  $\begin{array}{l}x=\frac{-\left(-i\right)\pm \sqrt{{\left(-i\right)}^2-4}}{2}\\ =\frac{i\pm \sqrt{i^2-4}}{2}\\ =\frac{i\pm \sqrt{-1-4}}{2}\\ =\frac{i\pm i\sqrt{5}}{2}\end{array}$

Hence, the roots of $z^2-iz+1=0are$

  $\frac{i+i\sqrt{5}}{2}and\frac{i-i\sqrt{5}}{2}$ .

De Moivre’s Theorem is not necessary.

Conclusion:

Hence, The roots of $z^2-iz+1=0$ are $\frac{i+i\sqrt{5}}{2}and\frac{i-i\sqrt{5}}{2}$ .

(c)

To determine

To solve: given equation.

Answer to Problem 100E

The roots of $z^2-\left(2-i\right)z-\frac{1}{4}i=0are$

  $x=\frac{\left(2-i\right)+\sqrt{3-3i}}{2}andx=\frac{\left(2-i\right)-\sqrt{3-3i}}{2}$ .

Explanation of Solution

Given:

  $z^2-\left(2-i\right)z-\frac{1}{4}i=0$

Calculation:

Consider $z^2-\left(2-i\right)z-\frac{1}{4}i=0$

Find the roots of $z^2-\left(2-i\right)z-\frac{1}{4}i=0$using the quadratic formula.

Use De Moivre’s Theorem to rewrite the roots. Recall that for a polynomial $a{x}^2+bx+c$ , the quadratic formula gives the roots:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In this case, $a=1,\text{ b}=-\left(2-i\right),and\text{ c}=-\frac{1}{4}i$ .

Thus, the quadratic formula gives:

  $\begin{array}{l}x=\frac{-\left(-\left(2-i\right)\right)\pm \sqrt{{\left(-\left(2-i\right)\right)}^2-4\left(-\frac{1}{4}i\right)}}{2}\\ x=\frac{\left(2-i\right)\pm \sqrt{{\left(2-i\right)}^2+i}}{2}\\ x=\frac{\left(2-i\right)\pm \sqrt{4-4i+i^2+i}}{2}\\ x=\frac{\left(2-i\right)\pm \sqrt{4-3i+i^2}}{2}\\ x=\frac{\left(2-i\right)\pm \sqrt{3-3i}}{2}\end{array}$

Hence, the roots of $z^2-\left(2-i\right)z-\frac{1}{4}i=0are$

  $x=\frac{\left(2-i\right)+\sqrt{3-3i}}{2}andx=\frac{\left(2-i\right)-\sqrt{3-3i}}{2}$ .

De Moivre’s Theorem is not necessary.

Conclusion:

Hence, the roots of $z^2-\left(2-i\right)z-\frac{1}{4}i=0are$

  $x=\frac{\left(2-i\right)+\sqrt{3-3i}}{2}andx=\frac{\left(2-i\right)-\sqrt{3-3i}}{2}$ .