Question

(i) Show that 1 +i and 1 - i satisfy the equation
$z {}^{2} - 2z + 2 = 0$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm :\longmapsto\: {z}^{2} - 2z + 2 = 0$

Since, its a quadratic equation, so its roots can be evaluated using Quadratic Formula.

We know,

Quadratic Formula is

$\rm :\longmapsto\:z = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

Here,

$\rm :\longmapsto\:a = 1$

$\rm :\longmapsto\:b = - 2$

$\rm :\longmapsto\:c= 2$

So, on substituting the values of a, b and c, we get

$\rm :\longmapsto\:z = \dfrac{ - ( - 2) \: \pm \: \sqrt{ {( - 2)}^{2} - 4(1)(2)} }{2(1)}$

$\rm :\longmapsto\:z = \dfrac{2 \: \pm \: \sqrt{{4 - 8 }}}{2}$

$\rm :\longmapsto\:z = \dfrac{2 \: \pm \: \sqrt{{- 4}}}{2}$

$\rm :\longmapsto\:z = \dfrac{2 \: \pm \: 2i}{2}$

$\bf\implies \:z = 1 \: \pm \: i$

Hence,

$\green{\bf :\longmapsto\:1 + i \: and \: 1 - i \: satisfy \: {z}^{2} - 2z + 2 = 0}$

Alternative Method

Given Quadratic Equation is

$\rm :\longmapsto\: {z}^{2} - 2z + 2 = 0$

can be rewritten as

$\rm :\longmapsto\: {z}^{2} - 2z + 1 + 1 = 0$

$\rm :\longmapsto\:( {z}^{2} - 2z + 1)+ 1 = 0$

$\rm :\longmapsto\:( {z}^{2} - 2z + {1}^{2} ) - ( - 1) = 0$

$\rm :\longmapsto\: {(z - 1)}^{2} - {i}^{2} = 0 \: \: \: \: \: \: \: \: \purple{\{\bf \because \: {i}^{2} = - 1 \}}$

$\rm :\longmapsto\:(z - 1 + i)(z - 1 - i) = 0$

$\red{\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\bf\implies \:z = 1 - i \: \: or \: \: 1 + i$

Hence,

$\red{\bf :\longmapsto\:1 + i \: and \: 1 - i \: satisfy \: {z}^{2} - 2z + 2 = 0}$