Question 187572
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ sqrt{-1}]


is not defined in the real numbers.  In order to deal with a radicand that is less than zero, a non-Real number, <i>i</i> (for imaginary), was defined as *[tex \Large i^2 = -1].  This number, along with any necessary real coefficients, comprise the Complex numbers.  Complex numbers are of the form *[tex \Large a + bi] where <i>i</i> is the imaginary number and <i>a</i> and <i>b</i> are Real numbers.


Any time you use the quadratic formula and the discriminant (the *[tex \Large b^2-4ac] under the radical) is less than zero, you need to factor out a -1, the square root of which is <i>i</i>, and then you can deal with the real number root of whatever is left under the radical.


The answer you gave, *[tex \Large 3 \pm 2i] are the roots of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 6x + 13 = 0]


I know this because I multiplied:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - (3 + 2i))(x - (3 - 2i)) = x^2 - (3 + 2i)x - (3 - 2i)x + (3 + 2i)(3 - 2i) = x^2 - 6x + 9 + 4]


So, using the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{-(-6) \pm sqrt{36-52}}{2}=\frac{6 \pm sqrt{-16}}{2} = \frac{6 \pm sqrt{(-1)(4)(4)}}{2} = \frac {6 \pm 4i}{2} = 3 \pm 2i]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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