Sarthaks Test
0 votes
3 views
ago in Complex Numbers by (18.0k points)

√i

Please log in or register to answer this question.

1 Answer

0 votes
ago by (17.2k points)

Let, (a + ib)2 = 0 + i

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

⇒ a2 + (bi)2 + 2abi = 0 + i

Since i2 = -1

⇒ a2 - b2 + 2abi = 0 + i

Now, separating real and complex parts, we get

⇒ a2 - b2 = 0 …………..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 = 0

⇒ 1 – 4b4 = 0

⇒ 4b2 = 1

Simplify and get the value of b2 , we get,

⇒ b2 = -\(\frac{1}{2}\) or b2\(\frac{1}{2}\)

As b is real no. so, b2 = 3

b = \(\frac{1}{\sqrt2}\) or b = -\(\frac{1}{\sqrt2}\)

Therefore , a = \(\frac{1}{\sqrt2}\) or a = -\(\frac{1}{\sqrt2}\)

Hence the square root of the complex no. is \(\frac{1}{\sqrt2}\) + \(\frac{1}{\sqrt2}\)i and - \(\frac{1}{\sqrt2}\) - \(\frac{1}{\sqrt2}\)i.

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...