find thw square root of 3+2i
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Answer:
2.236 is square root of 3+2i
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Step-by-step explanation:
- Let Sq root 3 + 2i = x + iy
- So squaring both sides we get
- 3 + 2i = (x + iy)^2
- 3 + 2i = x^2 + y^2i^2 + 2xyi
- 3 + 2i = x^2 – y^2 + 2xyi (since i^2 = -1)
- Equating the terms we get
- Now 3 = x^2 – y2 and 2i = 2xyi
- So x^2 – y^2 = 3, 2xy = 2
- Now (x^2 + y^2)^2 = (x^2 – y^2)^2 + (2xy)^2
- = 3^2 + 2^2
- = 9 + 4
- = 13
- So (x^2 + y^2) = √13
- Now we have both the equations
- So x^2 - y^2 = 3
- And x^2 + y^2 = √13
- Adding we get
- 2x^2 = 3 + √13
- Or x^2 = 3 + √ 13 / 2
- Or x = ± √ 3 + √ 13 / 2
- Now substituting for x we get
- So x^2 – y^2 = 3
- 3 + √ 13/2 – y^2 = 3
- 3 + √ 13 – 2y^2 / 2 = 3
- 3 + √ 13 – 2y^2 = 6
- Or y^2 = 3 – √ 13 / - 2
- Or y = ± 3 – √13 / -2
- Now we need to keep in x + yi form.
- So x + yi = 3 + √13/2 + ± 3 – √13 / - 2 i
- So x + yi = 3 + √13/2 + 3 - √13 /2i
- Reference link will be
- https://brainly.in/question/8789141
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