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Note, first, that
i2 = -1; also, that
(1 + i)2 = 1 + 2i - 1 = 2i; and, finally, that
[±½(√2) ]2 = ½.
Thus, [±½(√2)(1 + i)]2 = i.
Therefore, [±½ (√2)(i - 1)]2 = -i, and
±½ (√2)(i - 1) = √-i.
* * *
In case calculus is familiar, you may prefer to find the desired roots in this manner:
eiθ = cosθ + i sinθ; thus, when
θ = ½ π, eiθ = i; and, when
θ = ¼ π, eiθ = +√i.
Then, +√i = cos ¼ π + i sin ¼ π
= (½√2) + i ½√2
= (½√2) (1 + i).
This gives,
+√-i = (+√i)(+√-1)
= i(+√i)
= (½√2) i(1 + i)
= (½√2) (i - 1).
Therefore the two roots are:
±√-i = ±(½√2)(i - 1).
* * *
The first method is simpler, for proving the result; the second method, for finding it.
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Use a calculator (if you need) to find the principal square root. The second square root is the negative of the number.
How do you find the two square roots of 23?
Find the square root of 23. The second one is the same number with a negative sign.
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