FIND THE SQUARE ROOT OF COMPLEX NUMBER

Find the square root of complex number :

Here we are going to see how to find the square root of complex number.

Let a + ib be a complex number such that √a + ib  =  x + iy.

Let us look in to some example problems to understand the concept.

Example 1 :

Find the square root of the following 

7 - 24i

Solution :

Let √7 - 24i  =  x + iy, then

√(7 - 24i)  =  x + iy

Taking squares on both sides

7 - 24i  =  (x + iy)²

7 - 24i  =  x² + (iy)² + 2(x)(iy)

7 - 24i  =  x² - y² + i 2xy

x² - y²  =  7   ------(1)     2xy  =  -24  ------(2)

We have a formula,

(x² + y²)²  =  (x² - y²)² + 4x²y²

=  (x² - y²)² + (2xy)²

=  7² + (-24)²

 (x² + y²)²  =  49 + 576  ==>  625

 (x² + y²)  =  √625  ==>  25

x² + y²  =  25    ------(3)

(1)  +  (3) ==>   x² - y² + x² + y²  =  25 + 7

2x²  =  32

x²  =  16 ==>  x  =  ± 4

Hence the square root of the given complex number are

4 - 3i or (-4 + 3i)

Example 2 :

Find the square root of the following 

-15 - 8i

Solution :

Let √-15 - 8i  =  x + iy, then

√(-15 - 8i)  =  x + iy

Taking squares on both sides

-15 - 8i  =  (x + iy)²

-15 - 8i  =  x² + (iy)² + 2(x)(iy)

-15 - 8i  =  x² - y² + i 2xy

x² - y²  =  -15  ------(1)     2xy  =  -8  ------(2)

We have a formula,

(x² + y²)²  =  (x² - y²)² + 4x²y²

=  (x² - y²)² + (2xy)²

=  (-15)² + (-8)²

 (x² + y²)²  =  225 + 64  ==>  289

 (x² + y²)  =  √289  ==>  17

x² + y²  =  17    ------(3)

(1)  +  (3) ==>   x² - y² + x² + y²  =  -15 + 17 

2x²  =  2

x²  =  1 ==>  x  =  ± 1

Hence the square root of the given complex number are

1 - 4i or (-1 + 4i)

Example 3 :

Find the square root of the following 

-8 - 6i

Solution :

Let √-8 - 6i  =  x + iy, then

√(-8 - 6i)  =  x + iy

Taking squares on both sides

-8 - 6i  =  (x + iy)²

-8 - 6i  =  x² + (iy)² + 2(x)(iy)

-8 - 6i  =  x² - y² + i 2xy

x² - y²  =  -8  ------(1)     2xy  =  -6  ------(2)

We have a formula,

(x² + y²)²  =  (x² - y²)² + 4x²y²

=  (x² - y²)² + (2xy)²

=  (-8)² + (-6)²

 (x² + y²)²  =  64 + 36  ==>  100

 (x² + y²)  =  √100  ==>  10

x² + y²  =  10    ------(3)

(1)  +  (3) ==>   x² - y² + x² + y²  =  -8 + 10 

2x²  =  2

x²  =  1 ==>  x  =  ± 1

Hence the square root of the given complex number are

1 - 3i or (-1 + 3i)

Example 4 :

Find the square root of the following 

-3 + 4i

Solution :

Let √-3 + 4i  =  x + iy, then

√(-3 + 4i)  =  x + iy

Taking squares on both sides

-3 + 4i = (x + iy)²

-3 + 4i = x² + (iy)² + 2(x)(iy)

-3 + 4i = x² - y² + i 2xy

x² - y²  =  -3  ------(1)     2xy  =  4  ------(2)

We have a formula,

(x² + y²)²  =  (x² - y²)² + 4x²y²

=  (x² - y²)² + (2xy)²

=  (-3)² + (4)²

 (x² + y²)²  =  9 + 16  ==>  25

 (x² + y²)  =  √25 ==>  5

x² + y²  =  5    ------(3)

(1)  +  (3) ==>   x² - y² + x² + y²  =  -3 + 5

2x²  =  2

x²  =  1 ==>  x  =  ± 1

Hence the square root of the given complex number are

1 + 2i or (-1 - 2i)

Example 5 :

If [(1 - i)/(1 + i)]¹⁰⁰ = a + ib, then find ab.

Solution :

Given that, [(1 - i)/(1 + i)]¹⁰⁰

= [(1 - i)/(1 + i)]

To convert it as complex number, we have to multiply both numerator and denominator by the conjugate of the denominator.

= [(1 - i)/(1 + i)] [(1 - i)/(1 - i)]

= (1 - i)² / (1² - i²)

= (1² - 2(1)(i) + i²) / (1 - (-1))

= (1 - 2i - 1) / (1 + 1)

= - 2i / 2

[(1 - i)/(1 + i)] = -i

We have simplified and received the answer -i.

[(1 - i)/(1 + i)]¹⁰⁰ = (-i)¹⁰⁰

= ((-i)²)⁵⁰

= (i²)⁵⁰

= (-1)⁵⁰

= 1

1 = a + ib

By comparing the corresponding terms, we get

a = 1 and b = 0

ab = 1(0)

= 0

Example 6 :

If 1 + i is the root of the equation x² + ax + b where a and b ∈ then find the value of a + b.

Solution :

Since one of the root is in the form of complex number, then other root will be its conjugate.

The roots are 1 + i and 1 - i

Sum of roots = -b/a

From the given equation, sum of roots = -a

1 + i + 1 - i = -a

2 = -a

a = -2

Product of roots = c/a

From the given equation, product of roots = b

(1 + i) (1 - i) = b

1² - i² = b

b = 1 + 1

b = 2

The value of a + b = -2 + 2 ==> 0.

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