Solve, help, answer?

Question

What is:

1+ i (square root of 3) / 1 - i (square root of three)


the i's are multiplied by the square root of 3 ;)



Gracias

Answer 1

Completing the Square and The Square Root Method



http://www.ltcconline.net/greenl/Courses/152B/QuadraticsLineIneq/compsq.htm

Answer 2

The denominator is the problem here, but it's easily taken care of if you multiply it by what is called it's 'complex conjugate'.

The complex conjugate of [1 - i*sqrt(3)] is [1 + i*sqr(3)] as you'll discover if you check a relevant text.

Now if you are going to multiply the denominator by this, then to keep the expression happy, you'll have to also multiply the numerator by the same thing, so that, in effect, you will be multiplying the expression by 1, which doesn't change it, but does make it easier to work with.
So we have :

{[1 + i*sqrt(3)] / [1 - i*sqrt(3)]} * {[1 + i*sqrt(3)] / [1 + i*sqrt(3)]}

The numerator is [1 + i*sqrt(3)] / [1 + i*sqrt(3)]

= [1 + i*sqrt(3)] ^ 2
= 1 + 2i*sqrt(3) - 3 (note that i^2 = -1)
= -2 + 2i*sqrt(3)

The denominator is [1 - i*sqrt(3)] / [1 + i*sqrt(3)]

= 1 - (-3)
= 4

Now, numerator / denominator is :

[-2 + 2i*sqrt(3)] / 4

Dividing through by the common factor, 2, gives :

[-1 + i*sqrt(3)] / 2

or (-1/2) + [sqrt(3) / 2] * i , in standard form.

Answer 3

You cannot solve this because it is not an equation. What you can do is rationalize the denominator by multiplying by the conjugate. Multiply numerator and denominator of your fraction by 1 + i(sqrt3)

You will get [1+i(sqrt3)]^2divided by 4

You get he 4 because when you mult. the denominator be it's conjugate you get 1-3i^2. since i^2 is neg1 you now have 1+3 or 4

If your teacher wants you to simplify the numerator, you would have 1 +2i(sqrt3) +3i^2 Replace i^2 with -1, get 1+2i(sqrt3)-3 which equals -2 +2i(sqrt3)

You have that divided by 4, so you can factor the numerator, then divide out the common factor 2.

2[-1+i(sqrt3)]/4

[-1+i(sqrt3)]/2

Answer 4

1-i(square root of 3)

Answer 5

(1 + i√3)
------------- =
(1 - i√3)

(1 + i√3) ... (1 + i√3)
------------- X ------------ =
(1 - i√3) .... (1 + i√3)

1 + 2i√3 + 3i²
------------------ =
1 - 3i²

-2 + 2√3i
-------------- =
4

-½ + ½ √3i