Question

find the square root of complex number of 2-2√3i

Answer 1

$\sqrt{3}$

Answer 2

Answer:

√3-i is the square root of 2-2√3 i

 

Step-by-step explanation:

Consider the given complex number 2-2√3 i equate it to x+iy

x+iy=2-2√3 i

Square the above equation both sides

We get

$(x+i y)^{2}=(2-2 \sqrt{3} i)^{2}$

Apply $(a+b)^2$ formula

Then we get

Real part=$x^{2}-y^{2}$=2

Imaginary part=2xy=-2√3 i  

xy=-√3 i  

$y=-\frac{\sqrt{3} i}{x}$

Substitute the value of y in real part

$\begin{array}{l}{x^{2}-\left(-\frac{\sqrt{3} t}{x}\right)^{2}=2} \\ {x^{2}-\frac{3}{x^{2}}=2}\end{array}$

By taking LCM

$\begin{array}{l}{x^{4}-3=2 x^{2}} \\ {x^{4}-2 x^{2}-3=0}\end{array}$

By factorizing the above equation

We get the factors as $\left(x^{2}+1\right)\left(x^{2}-3\right)=0$

Here $\left(x^{2}+1\right)$  is not equal to 0

So $\left(x^{2}-3\right)$ = 0

$x=\pm \sqrt{3}$

If  

x=√3        x=-√3  

$y=-\frac{\sqrt{3}}{\sqrt{3}} \quad y=\frac{\sqrt{3}}{\sqrt{3}}$

=-1             =1

Therefore √3-i is the square root of 2-2√3 i