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what is the sum of the square roots of 16i?
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\[i \sqrt{16}= \pm 4i\]
i think @casanovaxx has done it incorrectly
answer should be (4*\[e^{ipi/4}\])
and off course both +/- sign
Okay, the possible answers given are -8, 0, 4\[\sqrt{2}\] +4\[\sqrt{2i}\], and 8i
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sorry that third one is , \[4\sqrt{2}+4\sqrt{2i}\]
8i? or the correction i added for the long one?
square roots of \(i\) are \(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\) and \(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\) and get 0
the 16 of course is unimportant, you multiply the above numbers by 4 when you add you still get zero
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