Find the square root of the following complex numbers:
i) -8 - 6i
ii) 7 + 24i
Answers
Answer:
Step-by-step explanation:
Method:
First the given complex number is written as a perfect square by using suitable algebraic identity.
Then easily we can easily get the square root.
1.
- 8 - 6i
= 1 - 9 - 2×1×3i
= 1² +(3i)² - 2×1×3i
= (1-3i)²
√-8-6i = ±(1-3i)
√-8-6i = 1 - 3i, -1+3i
2.
7 + 24i
= 16 - 9 + 2×4×3i
= 4² +(3i)² +2×4×3i
= (4+3i)²
√7+24i = ±(4+3i)
√7+24i = 4+3i, -4-3i
Answer:
i) 1 - 3i or -1 + 3i
ii)-4 - 3i or 4 + 3i
Step-by-step explanation:
Hi,
i) Let x + iy = √-8 -6i
Squaring on both sides, we get
x² - y² + i(2xy) = -8 -6i
Two complex numbers a + ib = c + id ⇔ a = c and b = d
Real parts should be equal and Imaginary parts should be
equal,
Hence, x² - y² = - 8 ----(1)and
2xy = -6
xy = -3
Substituting y = -3/x in (1), we get
x² - (-3/x)² = -8
x² - 9/x² = -8
Put t = x², we get
t - 9/t = -8
t² + 8t -9 = 0
t² + 9t -t -9 = 0
(t + 9)(t - 1) = 0
Hence t = -9 or t = 1
But t is square of real number , so it cannot be -9
Hence, t = 1
x² = 1
x = ± 1
y = -3/x
I x = 1 , y = -3
If x = -1, y = 3
Hence, square root is 1 - 3i or -1 + 3i
ii) Let x + iy = √7 + 24i
Squaring on both sides, we get
x² - y² + i(2xy) = 7 + 24i
Two complex numbers a + ib = c + id ⇔ a = c and b = d
Real parts should be equal and Imaginary parts should be
equal,
Hence, x² - y² = 7 ----(1)and
2xy = 24
xy = 12
Substituting y = 12/x in (1), we get
x² - (12/x)² = 7
x² - 144/x² = 7
Put t = x², we geet
t - 144/t = 7
t² - 7t -144 = 0
t² + 9t -16t -9 = 0
(t + 9)(t - 16) = 0
Hence t = -9 or t = 1
But t is square of real number , so it cannot be -9
Hence, t = 16
x² = 16
x = ± 4
y = 12/x
I x = 4 , y = 3
If x = -4, y = -3
Hence, square root is -4 - 3i or 4 + 3i
Hope, it helps !