The square root of -1 is i.
So what is the square root of i?
Can we try $\sqrt i= a+bi$?
Or
$(\sqrt i)^2=(a+bi)^2$
Or perhaps another approach towards the solution?
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$e^{i\pi/4}=\frac1{\sqrt2}(1+i)$. – Parcly Taxel Nov 06 '21 at 06:37
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5Does this answer your question? What is the square root of "$i$"? – Sathvik Nov 06 '21 at 06:38
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"The surd of -1 is = i" $;-;$ No, the square root*s* of $-1$ are $\pm i$. "Can we try i= a+bi?" $;-;$ You must mean $i=(a+bi)^2$ instead. (Note: the quotes are from the original question, which has been edited a few times since, not by the OP.) – dxiv Nov 06 '21 at 06:39
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Also see my post for general case here. – Ng Chung Tak Nov 06 '21 at 08:06
1 Answers
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You’re on the right track with your start, now that it’s been corrected.
Squaring gives
$i=a^2+2abi+b^2i^2$ $i=a^2+2abi-b^2$
You can then equate real and imaginary parts to get simultaneous equations for $a$ and $b$.
tomi
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