Answer:
Explanation:
First you assume some complex number of the form
![a + bi[\tex] is the square root of [tex]3 - 4i[\tex].</p><p>Then, by the definition, that number squared is 3 - 4i.</p><p>And you end up with the following equation:</p><p>[tex](a+bi)^2 = 3 - 4i\\a^2 + 2abi - b^2 = 3 - 4i\\(a^2 - b^2) + (2ab)i = 3 - 4i]()
Then you assume the real part of the left is equal to 3 and the complex part
![2abi[\tex] is equal to [tex]-4i[\tex].</p><p>You end up with a system of equations:</p><p>[tex]a^2 - b^2 = 3\\2ab = -4](https://img.qammunity.org/2020/formulas/mathematics/high-school/ohhn68ida4lv7euq8tl53zuhsq8bxhikf1.png)
Then you simplify the 2nd equation to
![ab = -2[\tex], then you rewrite b in terms of a [tex]b = (-2)/(a)[\tex].</p><p>You plug your new definition into the first equation and you end up with:</p><p>[tex]a^2 - ((-2)/(a))^2 = 3\\a^2 - (4)/(a^2) = 3]()
You multiply the whole equation by
![a^2[\tex] as it is not equal to 0.</p><p>[tex]a^4 - 4 = 3a^2\\a^4 - 3a^2 -4 = 0](https://img.qammunity.org/2020/formulas/mathematics/high-school/k3tb6nnakqmze6ctgwyqfxaytxdduf7mvb.png)
We let
![t = a^2[\tex] and we end up with:</p><p>[tex]t^2 -3t - 4 = 0\\t_(12) = (3 \pm √(9 - 4(1)(-4)) )/(2) = (3 \pm √(25))/(2) = (3 \pm 5)/(2)\\t_1 = 4\\t_2 = -1](https://img.qammunity.org/2020/formulas/mathematics/high-school/yj5ppw0lcuo25yqdawjv5lhbbx2j47dg7o.png)
We then go back to the definition of
![t[\tex]:</p><p>[tex]t = a^2\\a^2 = 4 \mid a^2 = -1](https://img.qammunity.org/2020/formulas/mathematics/high-school/z7jc14u85q24jboyvzcmc62odlrgkz5kxb.png)
But since a is a real number we only use the first result:

We then solve for
![b[\tex]:</p><p>[tex]ab = -2\\b_1 = (-2)/(a_1)\\b_2 = (-2)/(a_2)\\b_(12) = \pm 1](https://img.qammunity.org/2020/formulas/mathematics/high-school/kyzespb2qql3fkmmena48oipxyn0pe7v9a.png)
We then write the newly achieved complex number:
.
Use which equation you please to find the magnitude of:
- the magnitude.
And to find the phase/angle.
