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square root of complex number

Note by Abhimanyu Manu
4 years ago

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if you are asking about the square root of a complex number in standard form ie \(a+ib\) the method is simple.

Let \(\sqrt{a+ib} = x+iy \Rightarrow a+ib = (x+iy)^{2}\)

simplifying we get \(a+ib = x^{2}-y^{2}+2xyi \Rightarrow a = x^{2}-y^{2}\) and \(b = 2xy\) Snehdeep Arora · 4 years ago

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Convert your complex number into polar form, and remember that \( \sqrt{z} = z^{\frac{1}{2}} \) Ryan Carson · 4 years ago

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find sq. root of complex no.z=1-i Abhimanyu Manu · 4 years ago

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@Abhimanyu Manu The question is ambiguous. Which square root should we give? Michael Tang · 4 years ago

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@Abhimanyu Manu use De Moivre's theorem Tilak Patel · 4 years ago

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