x^2+2i = 0

Can anyone show me how to get the answer?

2. Originally Posted by stupid_kid
x^2+2i = 0

Can anyone show me how to get the answer?

or

x^2 = -2i

Let x = a + bi (a,b real numbers)

(a + bi)^2 = -2i

a^2 - b^2 + 2abi = -2i

2ab = -2 gives ab = -1 and a^2 - b^2 = 0

a = -1/b so (-1/b)^2 - b^2 = 0

1/b^2 - b^2 = 0

Multiply by b^2

1 - b^4 = 0

b^4 = 1 gives b = 1 or b = -1

b = 1 gives a = -1/1 = -1 giving solution x = -1 + i

b = -1 gives a = -1/-1 = 1 giving solution x = 1 - i

This methos should be used if you haven't done De Moivre's theorem

3. Originally Posted by stupid_kid
x^2+2i = 0

Can anyone show me how to get the answer?

I don't have time to do the thing properly, I'll just have to hope you know that:
e^{I*x} = cos(x) + I*sin(x) where I^2 = -1.

We know that
-I = e^{-I*(pi)/2}, so:

x^2 + 2I = 0

x^2 = -2I

x = sqrt{-2I} = sqrt{2}*sqrt[e^{-I*(pi)/2}]

x = sqrt{2}*[e^{-I*(pi)/2}]^{1/2} <-- Since sqrt(a) = a^{1/2}

x = sqrt{2}*e^{-I*(pi)/4}

I'll have to let you take the rest from here.

-Dan

4. Originally Posted by stupid_kid
x^2+2i = 0

Can anyone show me how to get the answer?

The solutions to the equation,

x^2 = -i

Are,

x= sqrt(2)/2 +/- sqrt(2)/2 i

(You can confirm this with de Moiver's theorem. But I just happen to know this fact.

Now,

x^2 = -2i

When you take the square root (or attempt to) you get the square root of -i, which is shown above. Times the sqrt(2).

Thus,

sqrt(2) * (sqrt(2)/2 +/- sqrt(2)/2) i) = 1+/- i