Question

find square root of 3-4i

Answer 1

this is Ur required solution

Answer 2

Answer:

Two roots are -2 + i and 2 - i

Step-by-step explanation:

We have to find the square root of 3 - 4i

$let\sqrt{3 - 4i}\ =\ x + iy$

Squaring both the sides we get,

$3 - 4i\ =\ (x + iy)^{2}$

$3 - 4i\ =\ x^{2}\ -\ y^{2}\ +\ 2xyi$

On comparing we get,

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

2xy = -4

xy = -2

$x = \frac{-2}{y}$

Putting the values in equation (i)

$\frac{4}{y^{2}}\ -\ y^{2}\ =\ 3$

Taking L.C.M we get

$\frac{4 - y^{4}}{y^{2}}\ =\ 3$

4 - y⁴ = 3y²

y⁴ + 3y² - 4 = 0

y⁴ + 4y² - y² - 4 = 0

y²(y² + 4) -1(y² + 4) = 0

(y² - 1)(y² + 4) = 0

y² = 1;

y = ± 1

Second value of y cannot be possible

If y = 1 ; x = -2

If y = -1 ; x = 2

Two roots are -2 + i and 2 - i