Find the square
root of (-8-6i)
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Answered by
1
Find the square root of 8 – 6i. First method. Let z2 = (x + yi)2 = 8 – 6i. \ (x2 – y2) + 2xyi = 8 – 6i. Compare real parts and imaginary parts,.
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Answered by
0
Let the square root be x+iy
⇒(x+iy)=√ (8−6i)
Taking square on both the side
⇒x^2-y^2+2xyi=−8−6i
By comparing we get.
⇒x
2
−y
2
=8 and 2xy=−6
⇒xy=−3
⇒y=
x
−3
⇒x
2
−(
x
−3
)
2
=−8
⇒x
2
−
x
2
9
=−8
⇒x
4
−9=−8x
2
⇒x
4
+8x
2
−9=0
⇒x
4
+9x
2
−x
2
−9=0
⇒x
2
(x
2
−1)+9(x
2
−1)=0
⇒(x
2
+9)(x
2
−1)=0
⇒x
2
=−9;x
2
=1
Since x is real we get x=±1
Taking x=±1 we get y=−3
⇒ square root is 1−3i
Taking x=−1 we get y=3
⇒ square root is −1+3i
Square root is ±(1−3i)
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