Question

find the square root of –6+8i​

Answer 1

Given ,

The complex number is -6 + 8i

Let ,

$\tt \sqrt{ - 6 + 8i} = a + ib \: - - \: (i)$

Squaring on both sides , we get

-6 + 8i = (a + ib)²

-6 + 8 = (a)² + (ib)² + 2aib

-6 + 8 = (a)² - (b)² + 2aib

Compare real parts and imaginary parts on both sides , we get

-6 = (a)² - (b)² --- (ii)

and

8 = 2ab

b = 8/2a --- (iii)

Put the value of b = 8√2a in eq (iii) , we get

-6 = (a)² - (8/2a)²

-6 = (a)² - 64/4(a)²

Taking LCM , we get

$\tt \implies - 6 = \frac{4 {(a)}^{2} - 64}{4 {(a)}^{2} }$

$\tt \implies - 24 {(a)}^{2} = 4 {(a)}^{2} - 64$

$\tt \implies 4 {(a)}^{2} + 24 {(a)}^{2} - 64 = 0$

$\tt \implies {(a)}^{4} + 6 {(a)}^{2} - 16 = 0$

$\tt \implies {(a)}^{4} + 8 {(a)}^{2} - 2 {(a)}^{2} - 16 = 0$

$\tt \implies {(a)}^{2} \{{(a)}^{2} + 8 \} - 2 \{ {(a)}^{2} + 8 \} = 0$

$\tt \implies {(a)}^{2} - 2 = 0 \: \: or \: \: {(a)}^{2} + 8 = 0$

$\tt \implies a = \pm\sqrt{2 } \: or \: a = \pm\sqrt{ - 8}$

Since , a is a real number

Therefore , the value of a is ± √2

Put the value of a = ± √2 in eq (iii) , we get

b = 8/2√2 or b = -8/2√2

b = 2√2 or b = -2√2

Put the value of a and b in eq (i) , we get

√2 + 2√2i or -√2 - 2√2i

Therefore , the square root of given complex number is

• √2 + 2√2i or -√2 - 2√2i

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Answer 2

Given to find square root,

$\longrightarrow z=-6+8i$

Compare it with $z=a+ib.$

Then,

• $a=-6$

• $b=8$

Modulus of $z$ is,

$\longrightarrow r=\sqrt{(-6)^2+8^2}$

$\longrightarrow r=10$

Solution 1:-

We have, if,

$\longrightarrow z=a+ib,$

then,

$\longrightarrow \sqrt z=\pm\left(\sqrt{\dfrac{r+a}{2}}+i\sqrt{\dfrac{r-a}{2}}\right)$

Then,

$\longrightarrow \sqrt z=\pm\left(\sqrt{\dfrac{10-6}{2}}+i\sqrt{\dfrac{10+6}{2}}\right)$

$\longrightarrow\underline{\underline{\sqrt z=\pm\sqrt2\left(1+2i\right)}}$

Solution 2:-

Let us convert our complex number in polar form.

$\longrightarrow z=10\left(-\dfrac{6}{10}+i\cdot\dfrac{8}{10}\right)$

$\longrightarrow z=10\left(-\dfrac{3}{5}+i\cdot\dfrac{4}{5}\right)$

Let $\theta=\arg(z)$ so that $\cos\theta=\dfrac{-3}{5}.$

We have, if,

$\longrightarrow z=r(\cos\theta+i\sin\theta),$

then,

$\longrightarrow \sqrt z=\left[r(\cos\theta+i\sin\theta)\right]^{\frac{1}{2}}$

By De Moivre's Theorem,

$\longrightarrow \sqrt z=\pm\sqrt r\left(\cos\left(\dfrac{\theta}{2}\right)+i\sin\left(\dfrac{\theta}{2}\right)\right)$

$\longrightarrow \sqrt z=\pm\sqrt r\left(\sqrt{\dfrac{1+\cos\theta}{2}}+i\sqrt{\dfrac{1-\cos\theta}{2}}\right)$

Then,

$\longrightarrow \sqrt z=\pm\sqrt{10}\left(\sqrt{\dfrac{1+\dfrac{-3}{5}}{2}}+i\sqrt{\dfrac{1-\dfrac{-3}{5}}{2}}\right)$

$\longrightarrow \sqrt z=\pm\sqrt{10}\left(\sqrt{\dfrac{2}{10}}+i\sqrt{\dfrac{8}{10}}\right)$

$\longrightarrow \sqrt z=\pm\left(\sqrt2+i\sqrt8\right)$

$\longrightarrow\underline{\underline{\sqrt z=\pm\sqrt2\left(1+2i\right)}}$