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Let, (a + ib)2 = 0 + 4i
Now using, (a + b)2 = a2 + b2 + 2ab
a2 + (bi)2 + 2abi = 0 + 4i
Since i2 = -1
a2 - b2 + 2abi = 0 + 4i
Now, separating real and complex parts, we get
a2 - b2 = 0 …………..eq.1
2ab =4…….. eq.2
a = ![]()
Now, using the value of a in eq.1, we get
– b2 = 0
4 – b4 = 0
b4 = 4
Simplify and get the value of b2 , we get,
b2 = -2 or b2 = 2
As b is real no. so, b2 = 2
b=
or b= ![]()
Therefore , a=
or a= -![]()
Hence the square root of the complex no. is
+
i and -
-
i.