Answer:
Step-by-step explanation:
First you assume some complex number of the form [tex]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[/tex]
Then you assume the real part of the left is equal to 3 and the complex part [tex]2abi[\tex] is equal to [tex]-4i[\tex].
You end up with a system of equations:
[tex]a^2 - b^2 = 3\\2ab = -4[/tex]
Then you simplify the 2nd equation to [tex]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[/tex]
You multiply the whole equation by [tex]a^2[\tex] as it is not equal to 0.
[tex]a^4 - 4 = 3a^2\\a^4 - 3a^2 -4 = 0[/tex]
We let [tex]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[/tex]
We then go back to the definition of [tex]t[\tex]:
[tex]t = a^2\\a^2 = 4 \mid a^2 = -1[/tex]
But since a is a real number we only use the first result:
[tex]a^2 = 4\\a_{12} = \pm 2[/tex]
We then solve for [tex]b[\tex]:
[tex]ab = -2\\b_1 = \frac{-2}{a_1}\\b_2 = \frac{-2}{a_2}\\b_{12} = \pm 1[/tex]
We then write the newly achieved complex number:
[tex]a_1 + b_1i = \sqrt{3-4i} \mid a_2 +b_2i = \sqrt{3-4i} \\2-i = \sqrt{3-4i} \mid -2 + i = \sqrt{3-4i}[/tex].
Use which equation you please to find the magnitude of:
[tex]|X| = \sqrt{2^2 + 1^2} = \sqrt{5}[/tex] - the magnitude.
And to find the phase/angle.
[tex]\theta = arcsin(\frac{b}{\sqrt{a^2+b^2} } ) = arcsin(\frac{1}{\sqrt{5}}) = 26.565^o[/tex]
