Question

find the square root of following complex number 1 + I

Answer 1

We want (a+bi)2=1−i(a+bi)2=1−i.

Let’s do it without angles and trig.

a2−b2+2abi=1−i⟹ a2−b2=1a2−b2+2abi=1−i⟹ a2−b2=1and 2ab=−12ab=−1.

So we have b2=14a2b2=14a2 which means we have:

a2−14a2=1⟹4a4–4a2–1=0a2−14a2=1⟹4a4–4a2–1=0

Thus a2 = 4±32√8 = 1±2√2a2 = 4±328 = 1±22

This gives us b2 = −1±2√2b2 = −1±22

We must choose the ‘++’ case not the ‘−−’ case otherwise bb is not real. And we have that 2ab=−12ab=−1 so the signs of aa and bb must be opposite.

Thus a=±12√2–√+1−−−−−−√b=∓12√2–√−1−−−−−−√a=±122+1b=∓122−1

Thus, finally we have:

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

hope it's helpful for you