What is the Demoivre Theorem for solving complex numbers?

[John O' Meara]

\sqrt{z}=\sqrt{r}(\cos \frac{\theta}{2} + \pi \ + \ \iota \sin \frac{\theta}{2} +\pi)\\ ...(i)
Obtain from equation (i) this equation \sqrt{z} \ = \ +/- [\sqrt{\frac{1}{2} (\abs{z} + x)} \ + \ \mbox{ sign y} \ \iota \sqrt{\frac{1}{2} ( \abs{z}+x)}]\\.
Where sign y =1 it y greater than or equal to 0, sign y =-1 if y < 0 and all squares of positive numbers are taken with positive sign. And where on the rhs z=abs{z}.
Equation (i) =- \sqrt{r}(\cos \frac{\theta}{2} \ + \ \iota \sin \frac{\theta}{2} )\\
=> z = r(cos^2 \frac{\theta}{2} \ + \ 2 \iota \cos \frac{\theta}{2} sin\frac{ \theta}{2}\ + \ \iota^2 sin^2 \frac{\theta}{2}) \\
= r( \frac{1}{2}(1 \ + \ cos \theta ) \ + \ \iota sin \theta \ + \ \frac{1}{2}(1 \ - \ cos \theta))\\
= \frac{1}{2}r \ + \ \frac{1}{2} r cos\theta \ + \ \iota^2 \frac{r}{2} - \frac{1}{2} r cos \theta\\
= \frac{1}{2} (\abs{z} \ + \ x) \ + \ \iota y \ - \ \frac{1}{2}r \ + \ \frac{1}{2} ( \abs{z} \ - \ x)\\
THat is as far as I get in it. As you can see it is wrong. Note on the rhs z=abs{z}. Thanks for the help.

 

[Avodyne]

Thre is a missing right parenthesis in your first expression. And the equation you are supposed to obtain has no equal sign and so is not an equation.

 

[John O' Meara]

The proper title is: Square root of complex numbers.

 

[Avodyne]

You want to show that the two right-hand sides are equal. So why are you taking the square? And then you replace i^2 with 1 instead of -1 when going from the 1st to 2nd line of your "z =" equation. You should just get z = x + iy at the end of that. But it's not what you want to do in the first place.

Your pi's in the first equation are also confusing. I assume that they are really part of the arguments of the sine and cosine, and that you just left out the parentheses. Also that they are optional; that's what would give you the plus-or-minus in the 2nd equation. It would help a lot if you would be precise in writing things down.

 

[Grufey]

First, I say sorry for my english, I will try write good.

You can prove that with a algebraic method:

We search a number z such a z^2=w, then, it's a system equations of form:

x^2-y^2=u
2xy=v

If we raise to square both equations, and add the second equation to the fisrt equation:

u^2+v^2=(x^2+y^2)^2

then we obtein a new system equation that we can solve easily:

\sqrt{u^2+v^2}=x^2+y^2
x^2-y^2=u

I leave the rest for you, it is the key, raise to square and ad the equations.

greeting from Grufey

 

[John O' Meara]

square root of complex numbers

I squared it to get rid of a minus sign in front of \sqrt{r} \\. The proper equation is: \sqrt{r}( \cos (\frac{\theta}{2} + \pi ) \ + \iota \sin( \frac{\theta}{2}+\pi)) \\. Thanks for reminding me about the need to be precise in maths.
Welcome to the PF Grufey.

 

[John O' Meara]

I cannot follow Grufey's method he seems to be using two functions u and v, yet the section this question is in says nothing about functions. Later sections talk about complex functions.

 

[HallsofIvy]

No, u and v are real numbers, the real and imaginary parts of z^2= (x+ iy)^2

 

[Grufey]

Sorry, I left define z=x+iy and w=u+iv

I recommended begin with z^2=w, I think that the problem is the notation, my z is my square root, I have a number w and I search a z such a z=\sqrt{w}, and you have z an you search w such a w=\sqrt{z}[\tex]. Sorry, I make a mistake with the notation.<br /> <br /> I hope that I can explain something.<br /> <br /> If you have some problem with complex analysis I recommend the book of Marsden Hoffman of complex analysis or ask in PF XD<br /> <br /> If you like I can do all. Do you like?

 

[John O' Meara]

Yes, Please do the question, as the book I use is not into giving examples, and I am short of examples.

 

[Grufey]

Ok, let's go:

\sqrt{z}=\sqrt{r}(\cos \frac{\theta}{2} + \pi \ + \ \iota \sin \frac{\theta}{2} +\pi)\\

What is a square of a number z?. it is a number w such a the product with himself is equal to z. Then the last equation we can write as:

\sqrt{z}=w

This is too general, because the second member of the equation is always a complex number that we define, w.
Now we can raise to square:

z=w^2

This is the equation that I said in other post. If z=z+iy and w=u+iv we have two equation, beacause, the fist is la equality of real parts, and the second equation is the equality of imaginary parts.

x+iy=(u+iv)(u+iv)

then:

u^2+v^2=x
2uv=y

we can raise to square both equations, and add the second equation to the first equation:

x^2+y^2=(u^2+v^2)^2

The new system is:

\sqrt{x^2+y^2}=u^2+v^2
u^2-v^2=x


u=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}
v=\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}

This two equation show 4 solutions, but the problem only have 2 solution, well, we can demostrate if we replace the solutions in the equation w^2=z that the solution to the problem are:

\sqrt{z}=w=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}+sign(v)\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}

 

[Grufey]

Oh, the last equation, also you can see as:

\sqrt{z}=w=\sqrt{\frac{x+\left|z\right|}{2}}+sign( v)\sqrt{\frac{-x+\left|z\right|}{2}}
or

\sqrt{z}=w=\sqrt{\frac{Re(z)+\left|z\right|}{2}}+sign( v)\sqrt{\frac{-Re(z)+\left|z\right|}{2}}

This is trivial, but I say because you can see any time

I only hope that you can bear my english, I have really problems to write and talk.

greeting from Grufey

 

[John O' Meara]

Thanks for the help. Do you know who is the publisher of Marsden Hoffman's book.

 

[Grufey]

This is the book, in the link you have all information:

https://www.amazon.com/dp/071672877X/?tag=pfamazon01-20

 

[ThienAn]

The problem you are trying to solve is called Demoivre Theorem, or you can solve it by applying Polar and Rectangular Cooordinates (a+bi)