Question

find square root of Z is equal to 5 - 12i​

Answer 1

Answer:

Z = 5 - 12i

Sol:

√(5 - 12i) = a + bi

squaring both the side

5 - 12i = a² + 2abi + b²i²

5 - 12i = (a² - b²) + 2ab .............{i² = -1}

Comparing real and imaginary parts

a² - b² = 5........1 ; 2abi = -12i

2ab = -12

Now, (a²+b²) = (a² - b²) + (2ab)²

= (5)² + (-12)²

= 25 + 144

= 169

a² + b² = √(169)

a²+ b² = 13.......................2

Adding 1 and 2

a² - b² = 5

+ a² + b² = 13

_____________

2a² = 18

a² = 9

a = 3,,

substituting a = 3 in , 2ab = -12

2ab = -12

2×3×b= -12

6b = -12

b = -2,,

√(5 - 12i) = a + ib

√(5 - 12i) = 3 + -2i)

√(5 - 12i) = 3 - 2i is Answer

Answer 2

Given:

A complex number 5-2i

To Find:

Square root of 5-12i

Solution:

We know that,

$\sf{i^{2}={(\sqrt{-1}})^{2}=-1}$

$\sf{(a+ib)^{2}=a^{2}+(ib)^{2}+2abi=a^{2}-b^{2}+2abi}$

Let  $\sf{\sqrt{5-12i}=x+iy}$

On squaring both the sides, we get

$\sf{{(\sqrt{5-12i}})^{2} =(x+iy)^{2}}$

$\sf{5-12i=x^{2}-y^{2}+2xyi}$

On comparing both the sides, we get

Constant Term

$\sf{x^{2}-y^{2}=5--------(1)}$

Coefficient of i

$\sf{2xy= -12}$

$\sf{y=\dfrac{-12}{2x}}$

$\sf{y=\dfrac{-6}{x}----------(2)}$

Now, on putting value of x in (1), we get

$\sf{x^{2}-\bigg(\dfrac{-6}{x}\bigg)^{2}=5}$

$\sf{x^{2}-\bigg(\dfrac{36}{x^{2}}\bigg)=5}$

$\sf{\dfrac{x^{4}-36}{x^{2}}=5}$

$\sf{x^{4}-36=5x^{2}}$

$\sf{x^{4}-5x^{2}-36=0}$

$\sf{x^{4}-9x^{2}+4x^{2} -36=0}$

$\sf{x^{2}(x^{2}-9)+4(x^{2} -9)=0}$

$\sf{(x^{2}+4)(x^{2}-9)=0}$

$\sf{x^{2}=9,-4}$

Since, x² can't be negative

So, x²≠ -4

x²= 9

x= ±3

When x=3, y= -2

x= -3, y= 2

So, required complex number is 3-2i and -3+2i.

Hence, square root of 5-12i is 3-2i or -3+2i.