Question

3. find the square root of 1+i​

Answer 1

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The square root of 1 - i is √(√(2)+1/2) - i√(√(2)-1/2)

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Answer 2

An alternative method to obtaining the square root of 1−i (Other answers have used Euler's form of the complex number to obtain the answer)

1−i−−−−√=a+bi

on squaring both sides we get

1−i=a2−b2+2abi

equating the real parts and imaginary parts on the LHS & RHS we get

a2−b2=1 & 2ab=−1

So we have, b=−12a

→b2=14a2

Which will now give us a2−14a2=1⟹4a4–4a2–1=0

Therefore a2 = 4±32√8 = 1±2√2

Hence, b2 = −1±2√2

We will neglect the ‘ − ’ case otherwise b will be not real. And we have 2ab=−1 so the signs of a and b must be opposite.

Thus a=±12√2–√+1−−−−−−√b=∓12√2–√−1−−−−−−√

Thus, finally we have —

1−i−−−−√=[±12√(2–√+1−−−−−−√−i2–√−1−−−−−−√)]

I hope it's help

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