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

  • #1
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]
 

Answers and Replies

  • #2
1,013
70
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
179
0
Thanks! I got to the quartic quite quickly in my attempt, but I got scared by it because it is a quartic :tongue:.
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
1,013
70
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
256
0
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
Gib Z
Homework Helper
3,346
5
There are only two solutions.
 
  • #8
Mentallic
Homework Helper
3,798
94
[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
240
2
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
256
0
True enough, my mistake.
 
  • #11
15
0
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
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,116
152
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
Galileo
Science Advisor
Homework Helper
1,989
6
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
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,116
152
Great minds think alike :smile:

I just realized this is an old thread, and was revived today because of the following question:
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.
 

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

  • Last Post
Replies
5
Views
961
  • Last Post
Replies
5
Views
30K
Replies
45
Views
2K
Replies
3
Views
3K
  • Last Post
Replies
2
Views
768
  • Last Post
Replies
20
Views
116K
Replies
2
Views
751
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
13
Views
29K
  • Last Post
Replies
14
Views
4K
Top