Name: Geometric View of Complex Number
Uploaded: 2019-07-19
Duration: 10 min 19 s
Description: Complex numbers extend the concept of the one-dimensional number line to the two-dimensional complex plane by using the horizontal axis. Watch the video to l...

Question

Find the square root of

- 2 + 2 \sqrt{- 3}

.

Accepted Answer

Let $\sqrt{- 2 + 2 \sqrt{- 3}} = \sqrt{- 2 + i 2 \sqrt{3}} = \pm (x + i y), x, y \in R$

Squaring on both sides, we get

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

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

$\Rightarrow - 2 + i 2 \sqrt{3} = x^{2} - y^{2} + 2 x i y [∵ i^{2} = - 1]$

Now equate real and imaginary parts on each side, we get

$\Rightarrow x^{2} - y^{2} = - 2 \to (1), 2 x y = 2 \sqrt{3} \to (2)$

Now, ${(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2}$

$\Rightarrow {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + {(2 x y)}^{2}$

$\Rightarrow {(x^{2} + y^{2})}^{2} = 4 + 12$

$\Rightarrow {(x^{2} + y)}^{2} = 16$

$\Rightarrow (x^{2} + y^{2}) = \pm 4$

As $x$ and $y$ are real numbers, so $x^{2} + y^{2}$ cannot be negative.

$∴ x^{2} + y^{2} = 4 \to (3)$

Now, add $(1)$ and $(3)$ , we get

$\Rightarrow 2 x^{2} = 2$

$\Rightarrow x^{2} = 1$

$\Rightarrow x = \pm 1$

From $(2)$ , we have $y = \frac{\sqrt{3}}{x} = \frac{\sqrt{3}}{\pm 1} = \pm \sqrt{3}$

$∴ \sqrt{- 2 + i 2 \sqrt{3}} = \pm (1 + \sqrt{3} i)$ .

Find the square root of $- 2 + 2 \sqrt{- 3}$ .

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Find the square root of -2+2-3.

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Find the square root of $- 2 + 2 \sqrt{- 3}$ .