Question

What is the magnitude and phase of X.


`X=√(3-4i)`

Answer 1

Explanation:

First you assume some complex number of the form

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`

Then you simplify the 2nd equation to

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`

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`

We then go back to the definition of
`t[\tex]:</p><p>[tex]t = a^2\\a^2 = 4 \mid a^2 = -1`

But since a is a real number we only use the first result:

`a^2 = 4\\a_(12) = \pm 2`

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`

We then write the newly achieved complex number:

`a_1 + b_1i = √(3-4i) \mid a_2 +b_2i = √(3-4i) \\2-i = √(3-4i) \mid -2 + i = √(3-4i)`.

Use which equation you please to find the magnitude of:

`|X| = √(2^2 + 1^2) = √(5)` - the magnitude.

And to find the phase/angle.

`\theta = arcsin((b)/(√(a^2+b^2) ) ) = arcsin((1)/(√(5))) = 26.565^o`