Math help. Complex numbers?

Question

 

write a quadratic equation with soultion i and -i

Answer 1

The  two solutions of the equation are i , -i

There fore this is a second degree equation

The quadratic equation form is ax² +bx + c = 0

We can find the a,b and c values in the equation

Substitute two solutions i,-i in the  quadratic equation : ax² + bx + c = 0

Substitute first solution i in the equation : ax² + bx + c = 0

There fore ai² + bi +c = 0

Substitute i² = -1 in the ai² + bi + c = 0

a(-1) + bi + c = 0

-a + bi + c =0 -(3)

Substitute second solution -i in the equation : ax² + bx + c = 0

There fore a(-i)² + b(-i) + c = 0

a(i)² - bi + c = 0

Substitute i² = -1 in the ai² - bi + c = 0

-a -bi + c = 0 -(4)

Add third and forth equations then we get a equation

-2a -0i + 2c = 0

-2a + 2c = 0

Add 2a to each side

2a -2a + 2c = 2a

0 + 2c = 2a

2c = 2a

Divide each side by 2

c = a

Substitute c = a in the third equation : -a + bi + c =0

There fore  -a + bi + a = 0

a - a + bi = 0

0 + bi = 0

bi = 0

But i = √(-1)

There fore b = 0

Substitute the value b = 0 and c = a in the quadratic equation :ax² + bx + c = 0

ax² +0x + a = 0

ax² +0 + a =0

ax² + a = 0

Recall : Distributive property  a(b + c) = ab + ac

a(x² + 1) = 0

Where a  is not equal to 0 because This have a two solutions

There fore this is a second degree equation.

There fore x² +1 = 0

x² - (-1) = 0

Substitute -1 = i² in the complex numbers

x² - i² = 0

or

(x - i)(x + i) = 0

The equation is x² + 1 = 0 or (x - i)(x + i) = 0.