Answers: 3
![a + bi[\tex] is the square root of [tex]3 - 4i[\tex].Then, by the definition, that number squared is 3 - 4i.And you end up with the following equation:[tex](a+bi)^2 = 3 - 4i\\a^2 + 2abi - b^2 = 3 - 4i\\(a^2 - b^2) + (2ab)i = 3 - 4i](/tpl/images/0502/8319/aeaf9.png)
![2abi[\tex] is equal to [tex]-4i[\tex].You end up with a system of equations:[tex]a^2 - b^2 = 3\\2ab = -4](/tpl/images/0502/8319/9c7e4.png)
![ab = -2[\tex], then you rewrite b in terms of a [tex]b = \frac{-2}{a}[\tex].You plug your new definition into the first equation and you end up with:[tex]a^2 - (\frac{-2}{a})^2 = 3\\a^2 - \frac{4}{a^2} = 3](/tpl/images/0502/8319/c86d8.png)
![a^2[\tex] as it is not equal to 0.[tex]a^4 - 4 = 3a^2\\a^4 - 3a^2 -4 = 0](/tpl/images/0502/8319/1c680.png)
![t = a^2[\tex] and we end up with:[tex]t^2 -3t - 4 = 0\\t_{12} = \frac{3 \pm \sqrt{9 - 4(1)(-4)} }{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}\\t_1 = 4\\t_2 = -1](/tpl/images/0502/8319/bef35.png)
![t[\tex]:[tex]t = a^2\\a^2 = 4 \mid a^2 = -1](/tpl/images/0502/8319/1c75a.png)
![b[\tex]:[tex]ab = -2\\b_1 = \frac{-2}{a_1}\\b_2 = \frac{-2}{a_2}\\b_{12} = \pm 1](/tpl/images/0502/8319/a5417.png)
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- the magnitude.
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