# Evaluate √1-i

4 views
ago

√1-i

ago by (17.2k points)

Let, (a + ib)2 = 1 - i

Now using, (a + b)2 = a2 + b2 + 2ab

⇒ a2 + (bi)2 + 2abi = 1 – i

Since i2 = -1

⇒ a2 - b2 + 2abi = 1 - i

Now, separating real and complex parts, we get

⇒ a2 - b2 = 1…………..eq.1

⇒ 2ab = -1…….. eq.2

⇒ a = -1/2b

Now, using the value of a in eq.1, we get

⇒ $(-\frac{1}{2b})^2$ – b2 = 1

⇒ 1 – 4b4 = 4b2

⇒ 4b4 + 4b2 -1 = 0

Simplify and get the value of b2, we get,

Hence the square root of the complex no. is