Name: Quadratic Equations Involving Complex Numbers
Uploaded: 2019-07-19
Duration: 8 min 29 s
Description: We already have solved many quadratic equations with real coefficients using the formula. How to solve a quadratic equation that has complex coefficients? Le...

Question

Find the square root of the complex number

12 - 5 i

.

Accepted Answer

Let $\sqrt{12 - 5 i} = \pm (x - i y), x, y \in R .$

Squaring both sides, we get

$12 - 5 i = {(x - i y)}^{2}$

$\Rightarrow 12 - 5 i = x^{2} + (i y)^{2} - 2 x i y$

$\Rightarrow 12 - 5 i = x^{2} - y^{2} - 2 x i y, [∵ i^{2} = - 1]$

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

$\Rightarrow x^{2} - y^{2} = 12 . . . (1)$ and $12 - 5 i$

We know that ${(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2}$

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

$\Rightarrow {(x^{2} + y^{2})}^{2} = 144 + 25$

$\Rightarrow {(x^{2} + y^{2})}^{2} = 169$

$\Rightarrow (x^{2} + y^{2}) = \pm 13 .$

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

$∴ x^{2} + y^{2} = 13 . . . (3)$

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

$\Rightarrow 2 x^{2} = 25$

$\Rightarrow x^{2} = \frac{25}{2}$

$\Rightarrow x = \pm \frac{5}{\sqrt{2}}$ .

From $(2)$ , we have $y = \frac{5}{2 x} = \frac{5}{\pm 2 \times \frac{5}{\sqrt{2}}} = \pm \frac{1}{\sqrt{2}} .$

$∴ \sqrt{12 - 5 i} = \pm \frac{1}{\sqrt{2}} (5 - i) .$

Find the square root of the complex number $12 - 5 i$ .

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Find the square root of the complex number 12-5i.

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Find the square root of the complex number $12 - 5 i$ .