Question

find the square root of i(i=√-1)

Answer 1

Compare real parts and imaginary parts,. x2 – y2 = 8 (1). 2xy = -6 (2). Now, consider the modulus: |z|2= |z2|. \ x2 + y2 = Ö(82 + 62) = 10 (3). Solving (1) and (3), we get x2 ...

Answer 2

Answer:

$\boxed{ \sqrt{i} = \frac{i + 1}{\sqrt{2}}}$

Step-by-step explanation:

Good question:

We know that:

$i = \sqrt{-1}$

$\implies i^2 = - 1$...............................(1)

$\bf{Lets\:find\:(i+1)^2\:\::-}$

$\implies ( i + 1 )^2 = (i + 1)(i + 1)$

$\implies (i + 1)^2 = i(i + 1) + 1(i+1)$

$\implies (i + 1)^2 = i^2 + i + i + 1$

$\implies (i + 1)^2 = -1 + 1 + 2 i$   [ From (1) ]

$\implies (i + 1)^2 = 2 i$

Taking square root both sides:-

$\implies i + 1 = \sqrt{2}\times\sqrt{i}$

$\implies \sqrt{i} = \frac{i + 1}{\sqrt{2}}$

Hope this helps you...............

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