# Square of complex numbers

1. Dec 7, 2007

### 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.

Last edited: Dec 8, 2007
2. Dec 7, 2007

### 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.

3. Dec 8, 2007

### John O' Meara

The proper title is: Square root of complex numbers.

4. Dec 8, 2007

### 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.

5. Dec 9, 2007

### 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

6. Dec 10, 2007

### 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.

7. Dec 12, 2007

### 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.

8. Dec 12, 2007

### HallsofIvy

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

9. Dec 12, 2007

### 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 $$w=\sqrt{z}[\tex]. Sorry, I make a mistake with the notation. I hope that I can explain something. If you have some problem with complex analysis I recommend the book of Marsden Hoffman of complex analysis or ask in PF XD If you like I can do all. Do you like? Last edited: Dec 12, 2007 10. Dec 13, 2007 ### John O' Meara Yes, Please do the question, as the book I use is not into giving examples, and I am short of examples. 11. Dec 14, 2007 ### Grufey Ok, let's go: [tex]\sqrt{z}=\sqrt{r}(\cos \frac{\theta}{2} + \pi \ + \ \iota \sin \frac{\theta}{2} +\pi)\\$$

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:

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

12. Dec 14, 2007

### 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

Last edited: Dec 14, 2007
13. Dec 15, 2007

### John O' Meara

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

14. Dec 15, 2007

### Grufey

15. Dec 19, 2007

### 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)