Question

root of 1+I / root of 1-i​

Answer 1

Step-by-step explanation:

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=1 and 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 aa and bb are not real. And we have that 2ab=−