Chapter 7.3, Problem 43WE

(a)

To determine

Find the roots of $i{z}^2-3z-2i=0$

Answer to Problem 43WE

-1

Explanation of Solution

Formula Used :

Discriminant of the equation $a{x}^2+bx+c=0\text{ is }\Delta =b^2-4ac$ .

Calculation:

Given equation is $i{z}^2-3z-2i=0$ .

Compare it with the standard quadratic equation , a = i , b = -3 , c = - 2i .

So, the discriminant is :

  $\begin{array}{l}\Rightarrow \Delta =b^2-4ac\\ \Rightarrow \Delta ={(}^{-}-4i(-2i)\\ \Rightarrow \Delta =9-\mathrm{8..................}[i^2=-1]\\ \Rightarrow \Delta =1\end{array}$

So, the discriminant is -1.

(b)

To determine

Find the roots of $i{z}^2-3z-2i=0$

Answer to Problem 43WE

The roots are $\alpha =-2i;\beta =-i$

Explanation of Solution

Formula Used :

Roots of the standard quadratic equation $a{x}^2+bx+c=0\text{ are }x=\frac{-b\pm \sqrt{\Delta }}{2a}$ , where $\Delta$ is the discriminant .

Calculation:

Given equation is $i{z}^2-3z-2i=0$ .

Compare it with the standard quadratic equation , a = i , b = -3 , c = - 2i .

From part (a) , the discriminant is -1.

So, the roots are :

  $\begin{array}{l}z=\frac{-b\pm \sqrt{\Delta }}{2a}\\ \Rightarrow z=\frac{-(-3)\pm \sqrt{1}}{2i}\\ \Rightarrow z=\frac{3\pm 1}{2i}\\ \Rightarrow z=\frac{3+1}{2i}\text{ and }z=\frac{3-1}{2i}\\ \Rightarrow z=\frac{4}{2i}\text{ and }z=\frac{2}{2i}\\ \Rightarrow z=\frac{2}{i}\text{=}\frac{2i}{i^2}\text{=}\frac{2i}{-1}\text{ and }z=\frac{i}{i^2}=\frac{i}{-1}\mathrm{......}[i^2=-1]\\ \Rightarrow \Rightarrow z=-2i\text{ and }z=-i\end{array}$

So, the roots are $\alpha =-2i;\beta =-i$

(c)

To determine

Explain why part (a) and (b) do not contradict the rule that quadratic with positive discriminant has real roots.

Explanation of Solution

Calculation:

From part (a) and (b) , we have

Given equation is $i{z}^2-3z-2i=0$ has discriminant 1 and roots are $\alpha =-2i;\beta =-i$ .

Since the equation is over complex numbers , that is , the variables comes from the field of complex numbers , hence here “real” means the roots are “complex” , since variables are complex.

Hence the rule that quadratic with positive discriminant has real roots has no contradiction .