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 = 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.
comparing real and imaginary parts ,we have
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 solutions of the equation are
(1+i) and (-1-i).
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