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find argument of question 5 + root 2 iota by 1 minus root 2iota

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Agam answered 2 month(s) ago

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

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

Class-XI Maths

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Asked by Adithyan

May 6

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Raghunath , SubjectMatterExpert

Member since Apr 11 2014

Let √(5 – 12i) = x + iy

Squaring both sides, we get

5 – 12i = x2 + 2ixy +(iy)2 = x2 – y2 + 2xyi.

Comparing real and imaginary parts , we get

5 = x2 – y2 ———– (1) and xy = – 6 ———— (2)

Squaring (1), we get

25 = (x2 – y2)2 = (x2 + y2)2 – 4x2y2

⇒ 25 = (x2 + y2)2 – 4(– 6)2

⇒ (x2 + y2)2 = 169

⇒ x2 + y2 = 13 ———- (3)

Adding (1) and (3) we get

2x2 = 18 

=> x = ± 3.

Subtracting (1) from (3) we get

2y2 = 8

=> y = ± 2.

Therefore,  √(5 – 12i) = 3 + 2i or, 3 – 2i or, – 3 + 2i or, – 3 – 2i.

As imaginary part of 5 – 12i is negative, the square root is  ±(3 – 2i)