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

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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 =...

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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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