Finding the square root of a+bi (complex number)

In summary, Roger Penrose discusses the square root of a complex number in terms of its real and imaginary parts. The solution involves using polar coordinates and taking the square root of the magnitude and half of the angle. There are two solutions, one of which is the negative of the other.
  • #1
Georgepowell
179
0
I was reading Roger Penrose' book "The Road to reality". He mentioned the square root of a+bi in terms of a and b. I am trying to figure his answer out for my self but am struggling. Here goes:[tex](x+yi)^2=a+bi[/tex]

[tex]x^2+2xyi-y^2=a+bi[/tex]

[tex]x^2-y^2=a[/tex]

[tex]2xy=b[/tex]

I can't rearrange these two equations to get x and y in terms of a and b. Even if I use a computer program to solve them for me, I get really complicated answers. Not like the solution in the book. Am I doing it wrong? Here is the solution he gives:

I have checked it and it works quite cleverly.

[tex]\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2}}[/tex]
 
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  • #2
Solve for x in the second equation to get x = b/(2y), which you can then plug into the first equation to get a quartic in y:
[tex]0 = y^4 + ay^2 + \left(\frac{b}{2}\right)^2[/tex]
Complete the square (using y2 as the variable in order to separate it from a) and solve for y2.
That's a great book. :)
 
  • #3
Thanks! I got to the quartic quite quickly in my attempt, but I got scared by it because it is a quartic :-p.
Although you wrote that slightly wrong. It should have been - (b/2)² (not plus). Here is the working for anyone that is interested.

[tex]0=y^4+ay^2-\frac{b^2}{4}[/tex]

[tex]0=(y^2+\frac{a}{2})^2-\frac{a^2}{4}-\frac{b^2}{4}[/tex]

[tex]y^2+\frac{a}{2}=\sqrt{\frac{1}{4}(a^2+b^2)}[/tex]

[tex]y=\sqrt{\frac{1}{2}(-a+\sqrt{(a^2+b^2)})}[/tex]

to get x

[tex]x^2-\frac{1}{2}(-a+\sqrt{(a^2+b^2)}=a[/tex]

[tex]x=\sqrt{\frac{1}{2}(a+\sqrt{(a^2+b^2)})}[/tex]
 
  • #4
Indeed. :) Good work.
 
  • #5
aren't you guys forgetting that the quadratic equation has a +/- option? That gives two solutions for x in the equation ax2 + bx + c = 0. Since 'x' itself is square, and we are given a quartic equation to solve, that's a total of 4 possible solutions for just the 'a' variable. The same goes for the 'b' variable, so, unless I'm missing some connection, 4 independent solutions of two variables each (a and b), gives a maximum total of 16 possible solutions just for the square root. If the +/- are connected for the two variables, that's a minimum of 4 possible solutions.

I think a single solution is too simplistic, and that there's more than meets the eye. It's worth dissecting further to find the different possible solutions.
 
  • #6
There are indeed many solutions. Although this thread has been quite computational so it might not be apparent. You will get one solution for each analytic branch of the natural log that you choose.
 
  • #7
There are only two solutions.
 
  • #8
[tex]a-\sqrt{a^2+b^2}<0[/tex] for all real a,b so that eliminates the possibility of the negative solution from the [itex]\pm[/itex] in:

[tex]x=\sqrt{\frac{1}{2}(a\pm\sqrt{(a^2+b^2)})}[/tex]

and

[tex]y=\sqrt{\frac{1}{2}(-a\pm\sqrt{(a^2+b^2)})}[/tex]

since x,y must be real numbers.

edit: except in the trivial case of b=0 but then we're dealing with real numbers so there's no point in all this :smile:
 
  • #9
Paul Czerner said:
aren't you guys forgetting that the quadratic equation has a +/- option? That gives two solutions for x in the equation ax2 + bx + c = 0. Since 'x' itself is square, and we are given a quartic equation to solve, that's a total of 4 possible solutions for just the 'a' variable. The same goes for the 'b' variable, so, unless I'm missing some connection, 4 independent solutions of two variables each (a and b), gives a maximum total of 16 possible solutions just for the square root. If the +/- are connected for the two variables, that's a minimum of 4 possible solutions.

I think a single solution is too simplistic, and that there's more than meets the eye. It's worth dissecting further to find the different possible solutions.

I think you are trying to argue with the fundumental theorem of algebra here,
Be careful
 
  • #10
True enough, my mistake.
 
  • #11
slider142 said:
Solve for x in the second equation to get x = b/(2y), which you can then plug into the first equation to get a quartic in y:
[tex]0 = y^4 + ay^2 + \left(\frac{b}{2}\right)^2[/tex]
Complete the square (using y2 as the variable in order to separate it from a) and solve for y2.
That's a great book. :)

Could someone please explain how the above equation was derived?
 
  • #12
Even if a and b are not real the "i" will cancel out giving two solutions which are the same.
 
  • #13
It's a lot easier to work this out in polar coordinates.

z = r e

√z = (r e)1/2

= r1/2 eiθ/2

(In other words, take the square root of the magnitude, and one half of the angle.)

Replacing θ with θ+2π does not change the value of z, and it gives the other root:

√z = r1/2 ei(π+θ/2)
 
  • #14
It is easier, in my view, to work from polar coordinates.
[itex]z=a+ib=|z|\exp(i\theta)=w^2[/itex].
So [itex]w=\sqrt{|z|}(\cos(\theta/2)+i\sin(\theta/2))[/itex].

Now use [itex]\cos(2\theta)=\sqrt{\frac{1}{2}(1+\cos \theta)}[/itex] and similar for the sine.
Since [itex]\cos(\theta)=a/\sqrt{a^2+b^2}[/itex] you can write down the expression. The other solution is just minus this number.

EDIT: Redbelly beat me to it :P
 
  • #15
Great minds think alike :smile:

I just realized this is an old thread, and was revived today because of the following question:
dink87522 said:
slider142 said:
Solve for x in the second equation to get x = b/(2y), which you can then plug into the first equation to get a quartic in y:
[tex]0 = y^4 + ay^2 + \left(\frac{b}{2}\right)^2[/tex]
Complete the square (using y2 as the variable in order to separate it from a) and solve for y2.
That's a great book. :)
Could someone please explain how the above equation was derived?

See post #1, where the OP got to the following two equations:
(1) x2 - y2 = a
(2) 2xy = b​
From (2), we know that x = b/2y
Substitute for x in (1) to get
b2/4y2 - y2 = a​
Multiply through by y2, and get all terms on the right side of the equation, to get
0 = y4 + ay2 - b2/4​
Note the "-" sign which was missed in the post you quoted, but corrected in Post #3.
 

FAQ: Finding the square root of a+bi (complex number)

1. What is the formula for finding the square root of a+bi?

The formula for finding the square root of a+bi is √(a+bi) = ±√(r) + [(a+bi) / (2√(r))], where r = √(a^2 + b^2).

2. How do you simplify the square root of a+bi?

To simplify the square root of a+bi, first find the value of r using the formula r = √(a^2 + b^2). Then, rewrite the complex number as √(r)[cos(θ) + i sin(θ)], where θ = tan^-1(b/a). Finally, take the square root of r and multiply it by the simplified expression inside the square root.

3. Can the square root of a complex number have multiple solutions?

Yes, the square root of a complex number can have multiple solutions, as seen in the formula √(a+bi) = ±√(r) + [(a+bi) / (2√(r))]. The ± symbol indicates that there are two possible solutions for the square root, one positive and one negative.

4. How do you find the principal square root of a+bi?

The principal square root of a+bi is the positive solution of the square root, which can be found by using the formula √(a+bi) = √(r) + [(a+bi) / (2√(r))]. This is the same as taking the positive value of the ± symbol in the general formula.

5. Are there any special cases when finding the square root of a+bi?

Yes, there are special cases when finding the square root of a+bi. If the complex number is purely imaginary (a=0), then the formula simplifies to √(bi) = ±√(b/2)i. If the complex number is purely real (b=0), then the formula simplifies to √(a) = ±√(a).

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