Complex numbers exercise

  • Thread starter renvox
  • Start date
  • #1
10
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Homework Statement


Hi,
I have to solve this exercise:
"Given that (a+bi)^2 = 3+4i obrain a pair of simultaneous equations involving a and b. Hende find the two square roots of 3+4i. Hence find the two suqre roots of 3+4i."
I don't really know to do do it.

2. The attempt at a solution
Well, that's what I thought:
(a+bi)^2 = 3+4
we expand:
a^2+2abi-b^2=3+4i
we group them:
2abi-4i=-a^2+b^2+3
common factor:
2i(ab-2)=-ab^2+b^2+3
2i=(-a^2+b^2+3)/(ab-2)

And now I don't even know if what I've done is good, and if it's good what should be the next step?
 

Answers and Replies

  • #2
867
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Starting at a2 + 2abi - b2 = 3 + 4i, you want to equate the real and imaginary parts.
For example, if c + di = x + yi, then c = x and d = y. This way you'll get a system of equations to solve.
 
  • #3
75
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Hi renvox.

Your initial equation [tex]\left(a+bi\right)^{2}=3+4i[/tex] involves the equality of two complex numbers. This implies that both the real part and the imaginary part of the two numbers must be equal.
 
  • #4
10
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Ok, so it will be:
2abi=4i
and
a^2-b^2=3
And now what do I do? :s
 
  • #5
75
0
You have a system of two equations and two variables...
 
  • #6
10
0
Sorry, it's my second day with complex numbers.
2abi=4i
a=sqrt(3+b^2)
therefore:
2bi*sqrt(3+b^2)=4i
-4bi*(3+b^2)=-16
-12b^2-4b^4=-16
dividing by -4
b^4+3b^2=4
common factor b^2
b^2(b^2+3)=4
right?
 
  • #7
75
0
The system of equations you must solve is the following:

From the equality of the real parts: [tex]a^{2}-b^{2}=3[/tex]
From the equality of the imaginary parts: [tex]2ab=4[/tex]
 
  • #8
10
0
lol, i'm making new laws of complex numbers :x
so:
b=2/a
a^2-2/a=3
a^3-3a-2=0
do I have to use cubic equation or is there another way to do it?
 
  • #9
867
0
Make sure you're doing your substitution correct. In a2 - b2 = 3, you have to square the 2/a when you substitute it in for b. So you'll get
[tex]a^2 - \left(\frac{2}{a}\right)^2 = 3[/tex]

Try taking it from there. :smile:
 
  • #10
10
0
thanks guys, you are awesome.
a^2-(4/a^2)=3
a^4 - 3^2 -4 = 0
by plugging: a=-2 or 2
so b= 1 or -1
thanks!
 
  • #11
eumyang
Homework Helper
1,347
10
thanks guys, you are awesome.
a^2-(4/a^2)=3
a^4 - 3^2 -4 = 0
Forgot an a2. This should be
[tex]a^4 - 3a^2 - 4 = 0[/tex]
by plugging: a=-2 or 2
By guess-and-check? You realize this is factorable, right?
[tex]a^4 - 3a^2 - 4 = (a^2 - 4)(a^2 + 1)[/tex]
 

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