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Note by Abhimanyu Manu 3 years, 1 month 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 · 3 years, 1 month ago

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Convert your complex number into polar form, and remember that \( \sqrt{z} = z^{\frac{1}{2}} \) – Ryan Carson · 3 years, 1 month ago

find sq. root of complex no.z=1-i – Abhimanyu Manu · 3 years, 1 month ago

@Abhimanyu Manu – The question is ambiguous. Which square root should we give? – Michael Tang · 3 years, 1 month ago

@Abhimanyu Manu – use De Moivre's theorem – Tilak Patel · 3 years, 1 month ago

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TopNewestif 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 · 3 years, 1 month ago

Log in to reply

Convert your complex number into polar form, and remember that \( \sqrt{z} = z^{\frac{1}{2}} \) – Ryan Carson · 3 years, 1 month ago

Log in to reply

find sq. root of complex no.z=1-i – Abhimanyu Manu · 3 years, 1 month ago

Log in to reply

– Michael Tang · 3 years, 1 month ago

The question is ambiguous. Which square root should we give?Log in to reply

– Tilak Patel · 3 years, 1 month ago

use De Moivre's theoremLog in to reply