Question

प्रश्न 04. सम्मिश्र संख्या 7+ 24 I
का वर्गमूल ज्ञात कीजिए ।
Find the square root of the complex number 7+24 i​

Answer 1

Given :  7+24 i​

To Find : square  root

Solution :

Let say square root = a + ib

a + ib = √(7 + 24 i)

Squaring both sides

=> a²  + (ib)² + 2abi  = 7 + 24i

=> a² - b²  + 2abi  = 7 + 24 i

2ab = 24

=> ab = 12

12 = 1 * 12  = 2 * 6 = 3 * 4

a² - b² = 7

=> (a + b)(a - b) = 7

a = 4 and b = 3 will satisfy this

4 + 3i   is square root   7+ 24 i

or other way

(a² + b²)²  = (a² - b²)² + 4a²b²

=> (a² + b²)²  = (7)² + 24²         4a²b²= (2ab)²

=> a² + b² = 25

a² + b² = 25

a² - b² = 7

=> a²  = 16  and b²  = 9

=> a = ± 4  and b = ±3

4 + 3i  and -4 - 3i     is square root   7+ 24 i

Learn More:

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Answer 2

Answer:

$\bf \underline{ \: solution} \: \\ \\ \\ \bf \: we \: have \: \\ \\ \\ \bf \sqrt{7 + 24i = {x}^{2} } - {y}^{2} + 2xyi \\ \\ \\ \bf \: {x}^{2} - {y}^{2} = 7 \: \: \: {x}^{2} - \frac{144}{ {x}^{2} } = 7 \\ \\ \\ \bf {x}^{4} - {7x}^{2} - 144 = 0 \\ \\ \\ \bf \: {x}^{2} ( {x }^{2} - 16) + a( {x}^{2} - 16) = 0 \\ \\ \\ \bf \: x = \pm \: y \\ \\ \\ \therefore \bf \: x = \frac{12}{x} \\ \\ \\ \bf \: y = 3 \: \: \: \: - 3 \\ \\ \\ \bf \therefore x = 4 \: \: \: y = 3 \\ \\ \\ \bf \: \: x = - 4 \: y = - 3 \\ \\ \\ \\ \bf \: \sqrt{7 + 24i} = \pm4 + 3i \: \\ \\ \\ \bf \: hence \: is \: the \: answer$