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Grade: 11
`        find the square root of the complex number:(3+4i)/(3-4i)`
10 months ago

## Answers : (1)

Arun
22559 Points
```							Dear studentÂ I am thinking that you are asking square root of any one.Â Set the square root of this equal to a + bi. Square both sides. Simplify. Then compare the real and imaginary parts separately and solve for a and b.Â âˆš(3 + 4i) = a + biÂ 3 + 4i = (a+bi)^2Â 3 + 4i = a^2 + 2abi + (bi)^2Â 3 + 4i = a^2 + 2abi - b^2Â So:Â 3 = a^2 - b^2, and 4i = 2abi, which meansÂ 3 = (a^2 - b^2) and 2 = abÂ Replace a = 2/b in the first equation to getÂ 3 = (2/b)^2 - b^2Â 3 = (4 / b^2) - b^2Â 3b^2 = 4 - b^4Â b^4 + 3b^2 - 4 = 0Â (b^2 + 4)(b^2 - 1) = 0Â So b^2 = -4, 1. We're only looking for real values of b, so b = Â±1. This means a = 2/b = Â±2. So 2 + i and -2 - i are two solutions.Â
```
10 months ago
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