Question

Evaluate

$\sqrt{7 - 24i}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\: \sqrt{7 - 24i}$

Let assume that

$\rm :\longmapsto\: \sqrt{7 - 24i} = x + iy - - - - (1)$

On squaring both sides, we get

$\rm :\longmapsto\:7 - 24i = {(x + iy)}^{2}$

$\rm :\longmapsto\:7 - 24i = {x}^{2} + {i}^{2} {y}^{2} + 2xyi$

$\rm :\longmapsto\:7 - 24i = {x}^{2} - {y}^{2} + 2xyi \: \: \: \: \: \: \: \{ \because {i}^{2} = - 1 \}$

So, on comparing real and Imaginary parts, we get

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: {x}^{2} - {y}^{2} = 7 \: }}} - - - - (2)$

and

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: 2xy = - 24 \: }}} - - - - (3)$

Now, we know that,

$\red{\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{ {( {x}^{2} - {y}^{2}) }^{2} + 4 {x}^{2} {y}^{2} } }$

can be further rewritten as

$\red{\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{ {( {x}^{2} - {y}^{2}) }^{2} + {(2xy)}^{2}}}$

On substituting the values from equation (2) and (3), we get

$\rm :\longmapsto\: {x}^{2}+{y}^{2} = \sqrt{ {( - 7)}^{2} + {(24)}^{2} }$

$\rm :\longmapsto\: {x}^{2}+{y}^{2} = \sqrt{ 49 + 576}$

$\rm :\longmapsto\: {x}^{2}+{y}^{2} = \sqrt{ 625}$

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: {x}^{2} + {y}^{2} = 25 \: }}} - - - - (4)$

On adding equation (2) and (4) we get

$\rm :\longmapsto\: {2x}^{2} = 32$

$\rm :\longmapsto\: {x}^{2} = 16$

$\rm \implies\:\boxed{ \tt{ \: x \: = \: \pm \: 4 \: }} - - - - (5)$

On Subtracting equation (2) from equation (4), we get

$\rm :\longmapsto\: {2y}^{2} = 18$

$\rm :\longmapsto\: {y}^{2} = 9$

$\rm \implies\:\boxed{ \tt{ \: y \: = \: \pm \: 3 \: }} - - - - (6)$

As from equation (3), we have

$\red{\rm :\longmapsto\:xy < 0 \: }$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 4 & \sf - 3 \\ \\ \sf - 4 & \sf 3 \end{array}} \\ \end{gathered}$

Hence,

$\purple{\bf :\longmapsto\: \sqrt{7 - 24i} = - 4 + 3i \: \: or \: \: 4 - 3i}$

Answer 2

Calculation :

♦ Let, √7 - 24 i = a + bi where a,b € R

Squaring on both sides we get,

7 + 24 i = (a + bi )²

7 + 24i = a² + b² i² + 2abi

7 + 24i = ( a² - b² ) + 2abi. ( i² = -1 )

Equating real and imaginary parts we get,

${a}^{2} - {b}^{2} = 7 \: \: and \: \: 2ab = 24 \\ {a}^{2} - {b}^{2} = 7 \: and \: \: b = \frac{12}{a} \\ {a}^{2} - ( \frac{12}{a} ) {}^{2} = 7 \\ {a}^{2} - \frac{144}{ {a}^{2} } = 7 \\ {a}^{4} - 144 - 7 {a}^{2}$

a⁴ - 7a² - 144 = 0

( a²- 16 ) (a² + 9 ) = 0

a² = 16. or a² = -9

But a € R

a² ≠ -9

therefore, a² = 16

a = +4 or -4

$when \: a = 4 \: \: \: b = \frac{12}{4} = 3 \\ when \: a = - 4 \: \: \: b = \frac{12}{ - 4} = - 3$

therefore, √ 7+ 24i = +( 4 + 3i ) or - ( 4 + 3i )