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`        find the square root of the complex number:(3+4i)/(3-4i)`
one year ago

```							Dear student I am thinking that you are asking square root of any one. 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. √(3 + 4i) = a + bi 3 + 4i = (a+bi)^2 3 + 4i = a^2 + 2abi + (bi)^2 3 + 4i = a^2 + 2abi - b^2 So: 3 = a^2 - b^2, and 4i = 2abi, which means 3 = (a^2 - b^2) and 2 = ab Replace a = 2/b in the first equation to get 3 = (2/b)^2 - b^2 3 = (4 / b^2) - b^2 3b^2 = 4 - b^4 b^4 + 3b^2 - 4 = 0 (b^2 + 4)(b^2 - 1) = 0 So b^2 = -4, 1. We're only looking for real values of b, so b = ±1. This means a = 2/b = ±2. So 2 + i and -2 - i are two solutions.
```
one year ago
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