Find the square root of 16-30i
Answers
Answered by
111
Hi ,
Let √( 16 - 30i ) = x - iy ---( 1 )
Do the square of equation ( 1 ) , we
get ,
16 - 30i = ( x - iy )²
= x² + ( iy )² - 2 × x × iy
= ( x² - y² ) - 2xyi
Compare both sides ,
x² - y² = 16 ---( 2 )
2xy = 30
y = 30/2x
y = 15/x ---( 3 )
Substitute y value in equation ( 2 ),
we get
x² - ( 15/x )² = 16
x² - 225/x² = 16
x⁴ - 225 = 16x²
x⁴ - 16x² - 225 = 0
x⁴ - 25x² + 9x² - 225 = 0
x² ( x² - 25 ) + 9 ( x² - 25 ) = 0
( x² - 25 ) ( x² + 9 ) = 0
x² - 25 = 0 or x² + 9 = 0
x² = 25
x = ± 5
Substitute x = ± in equation ( 3 ) ,
We get ,
y = 15/( ± 5 )
y = ± 3
Therefore ,
x = ± 5 , y = ± 3
√( 16 - 30 i ) = x - iy
= ± ( 5 - 3i )
I hope this helps you.
: )
Let √( 16 - 30i ) = x - iy ---( 1 )
Do the square of equation ( 1 ) , we
get ,
16 - 30i = ( x - iy )²
= x² + ( iy )² - 2 × x × iy
= ( x² - y² ) - 2xyi
Compare both sides ,
x² - y² = 16 ---( 2 )
2xy = 30
y = 30/2x
y = 15/x ---( 3 )
Substitute y value in equation ( 2 ),
we get
x² - ( 15/x )² = 16
x² - 225/x² = 16
x⁴ - 225 = 16x²
x⁴ - 16x² - 225 = 0
x⁴ - 25x² + 9x² - 225 = 0
x² ( x² - 25 ) + 9 ( x² - 25 ) = 0
( x² - 25 ) ( x² + 9 ) = 0
x² - 25 = 0 or x² + 9 = 0
x² = 25
x = ± 5
Substitute x = ± in equation ( 3 ) ,
We get ,
y = 15/( ± 5 )
y = ± 3
Therefore ,
x = ± 5 , y = ± 3
√( 16 - 30 i ) = x - iy
= ± ( 5 - 3i )
I hope this helps you.
: )
Answered by
16
Assumption
√16 - 30i = (x + iy) ..... (1)
(16 - 30i) = (x - iy)²
(16 - 30i) = (x² - y²) - i(2xy) ... (2)
Here
(x² - y²) = 16
2xy = 30
(x² + y²) = √{(x² - y²)² + 4x²y²}
= √(16² + 30²)
= √(256 + 900)
= √1156
= 34
Hence,
(x² - y²) = 16 .... (3)
Also,
(x² + y²) = 34 ..... (4)
x² = 25
x = √25
x = ±5
Also,
y² = 9
y = √9
y = ±3
Here
xy > 0
Therefore,
(x = 5 and y = 3) or (x = -5 and y = -3)
Hence,
√16 - 30i = (5 - 3i) or (-5 + 3i)
Similar questions