find the square root of
-5+i12
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Heya user,
Let 12i - 5 = [ a + ib ]² ------> where (a,b) are real no.s
=> 12i - 5 = ( a² - b² ) + 2abi
Comparing the parts,
( a² - b² ) = -5 and 2ab = 12
=> a = 6/b ;
Putting this in the other eqn.
( 36/b² - b² ) = -5
.'. we get two real soln.s for b---> b = 3 or b = -3
=> a = 2 or -2
Now, since ab = 6, signs of both a and b are same
Hence, 12i - 5 = [ 2 + 3i ]² or [ -2 - 3i ]²
.'. sqrt [ 12i - 5 ] = [ 2 + 3i ] or [ -2 - 3i ]
Let 12i - 5 = [ a + ib ]² ------> where (a,b) are real no.s
=> 12i - 5 = ( a² - b² ) + 2abi
Comparing the parts,
( a² - b² ) = -5 and 2ab = 12
=> a = 6/b ;
Putting this in the other eqn.
( 36/b² - b² ) = -5
.'. we get two real soln.s for b---> b = 3 or b = -3
=> a = 2 or -2
Now, since ab = 6, signs of both a and b are same
Hence, 12i - 5 = [ 2 + 3i ]² or [ -2 - 3i ]²
.'. sqrt [ 12i - 5 ] = [ 2 + 3i ] or [ -2 - 3i ]
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