# Thread: Compex Number Equation - Need Help

1. ## Compex Number Equation - Need Help

I have been reading A Long Way From Euclid a book by Constance Reid. In the chapter on complex numbers she gives an equation, and the solutions to the equation, but does not show how the solutions are arrived at. I am stumped, and (finally) realize I need help. Just as important as knowing how to solve this equation where i equals "1" i, I would like to know how to solve it if the i equals 2i, 3i, 4i. . .ni, where n is any real number. Can the equation be solved algebraically by converting the x and the i into a + bi form? Or, does DeMoivre's theorem for finding complex roots have to be used? Both? Or, somthing else?

The equation is

$x^2-i=0$

The solutions that Reid gives are --

$x=\pm\left(\frac{1+i}{\sqrt{2}}\right) \;=\; \sqrt{i}$

2. in complex polar form, $i = e^{i \frac{\pi}{2}} = \cos (\frac{\pi}{2}) + i \sin (\frac{\pi}{2})$

so $\sqrt{i} = e^{i\frac{\pi}{4}} = \cos (\frac{\pi}{4}) + i \sin (\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}$

3. Originally Posted by DeanSchlarbaum
... Constance Reid. In the chapter on complex numbers gives an equation, and the solutions to the equation, but does not show how the solutions are arrived at.
The equation is

$x^2-i=0$

The solutions that Reid gives are --

$x=\pm\left(\frac{1+i}{\sqrt{2}}\right) \;=\; \sqrt{i}$
I think she just the formula for a second degree equation.

4. Originally Posted by gciriani
I think she just the formula for a second degree equation.
gciriani: I don't quite understand your post. Could you explain in more detail, please.

5. Originally Posted by Random Variable
in complex polar form, $i = e^{i \frac{\pi}{2}} = \cos (\frac{\pi}{2}) + i \sin (\frac{\pi}{2})$

so $\sqrt{i} = e^{i\frac{\pi}{4}} = \cos (\frac{\pi}{4}) + i \sin (\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}$
Random Variable: You are using DeMoivre's Formula, if I'm not mistaken. Is that correct? If you care to, would you mind "walking me thougth the steps" to arrive at the solutions? Its nice to see the answer, but I still don't undertand how exactly you arrived at it. For instance, Reid says that one of the solutions is

$x=\left(\frac{-1-i} {-\sqrt{2}}\right)$

I don't see that in your single solution. What am I missing?

Also, you didn't address my question as to if the equation could be solved algrebraically by converting the $x^2$ and the
$i$ into the $a+bi$ form. Someone (not at MHF) told me it was possible, but didn't explain or show how. Do you have any input about this?

6. Originally Posted by DeanSchlarbaum
gciriani: I don't quite understand your post. Could you explain in more detail, please.
Sorry, my words fell through the keyboard's cracks.
She used the formula for a second degree equation x^2=+/-sqrt(i)
Also without invoking de Moivre, you can use the multiplication property for complex numbers: you add the angles and you multiply the radii. For this equation therefore you have to find the number that multiplied by itself gives i, i.e. you double the angle and you square the radius. For i the radius is 1 the angle is 90. Therefore the solution is a complex number with radius 1 and an angle half that: 45 degrees or -135 degrees. These are the two points given by the solution.

7. Please help me to find out,A problem on De Moivres theorem

First i apologize if i have posted at wrong place....

The problem is here:

Solve x^8+x^5+x^3+1 by using De Moivres theorem

An example in my book might help(didnt helped me at at all)

x^4-x^3+x^2-x+1

Sol:

The equation can be written as

x^4-x^3+x^2-x+1 = (x^5+1)/(x+1)
Hence the required roots of x^5+1 = 0 are same as those of (x^4-x^3+x^2-x+1)(x^5+1)

The equation (x^5+1) gives x^5 = -1 or x = (-1)^1/5

(......and so on)

Now what my trouble is how i can find equation like (x^5+1) for my problem stated at start of post?

I am newbie and poor in maths so little more explanation will help lot