# Math help. Complex numbers?

write a quadratic equation with soultion i and -i

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

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.