# Solve for complex z equation z^2=2i?

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### 2 Answers

You need to consider the following complex number `z = x + i*y` and you need to replace it in the given equation, such that:

`(x + i*y)^2 = 2i => x^2 + 2i*x*y + (i^2*y^2) = 2i`

Replacing `-1` for ` i^2` yields:

`x^2 - y^2 + 2i*x*y = 2i`

You need to equate the real parts both sides, such that:

`x^2 - y^2 = 0 => x^2 = y^2`

You need to equate the imaginary parts both sides, such that:

`xy = 1`

Squaring both sides yields:

`x^2*y^2 = 1`

Replacing `x^2` for `y^2` yields:

`x^2*x^2 = 1 => x^4 = 1 => x^4 - 1 = 0 `

Converting the difference of squares into a product yields:

`(x^2 - 1)(x^2 + 1) = 0`

Using the zero product rule, yields:

`x^2 - 1 = 0 => x^2 = 1 => x_(1,2) = +-1 => y = +-1`

`x^2 + 1 = 0 => x^2 = -1 => x_(1,2) = +-i => y = +-i`

**Hence, evaluating the solutions to the given equation yields `z = 1 + i , z = -1 - i, z = -1+i ` and **`z = 1 - i .`

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The last two are extraneous solutions. Only `+-(1+i)` are solutions.

`z^2=2i`

Let

Z=a+ib

Thus

`(a+ib)^2=2i`

`a^2-b^2+2iab=2i`

comparing real and imaginary parts ,we have

`a^2-b^2=0`

`2ab=2`

`ab=1`

But

`(a^2+b^2)^2=(a^2-b^2)^2+4(ab)^2`

`(a^2+b^2)^2=4`

`a^2+b^2=sqrt(4)=2`

since a and b are real numbers . Also `a^2 ,b^2>=0` therefore

`a^2+b^2>=0` .It ruled out the possibilitt of negative root.

Thus

`a^2-b^2=0`

`a^2+b^2=2`

`=>2a^2=2`

`a=+-1`

`and b=+-1`

`Thus`

`z=-1-i`

`and `

`z=1+i`

Thus solutions of the equation are

(1+i) and (-1-i).