Finding (2+6i)^(1/3): Proving a+bi Assumption is Correct

[Outrageous]

Find (2+6i)^(1/3). Then we need to let z=a+bi, then z^3... Then we will get a=? And b=?
Why we can do z=a+bi assumption? Any prove show what we assume is correct and how do we know the answer come out is correct?

 

[Matt Benesi]

http://en.wikipedia.org/wiki/Complex_logarithm

and:

http://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers

 

[Outrageous]

Sorry, can you tell what is the subtopic of my question in the reference given. Thanks

 

[HallsofIvy]

Find (2+6i)^(1/3). Then we need to let z=a+bi, then z^3... Then we will get a=? And b=?
Why we can do z=a+bi assumption? Any prove show what we assume is correct and how do we know the answer come out is correct?

I'm not sure what you mean by "z= a+ bi assumption". Typically that is NOT an "assumption", it is one way of defining complex numbers.

A more formal definition, from which you can then derive "z= a+ bi" is this:
The complex numbers is the set of all pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d), and with multipication defined by (a, b)(c, d)= (ac- bd, bc+ ad). One can then show that the usual "properties" of arithmetic (commutativity, associativity, etc.) hold. One can also show that the subset of all pairs in which the second member is 0 is "isomophic" to (identical to) the set of real numbers: (a, 0)+ (b, 0)= (a+ b, 0+ 0)= (a+ b, 0) and (a, 0)(b, 0)= (ab- 0(0), a(0)+ 0(b))= (ab, 0). We can "identify" complex numbers of the form (a, 0) with the real number, a.

One can also show that (0, 1)^2= -1: (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+1(0))= (-1, 0) which, as above, we identify with the real number -1. We define "i= (0, 1)" so that i^2= -1. And, then, we can write (a, b)= (a, 0)+ (0, b)= (a, 0)+ (b, 0)(0, 1)= a+ bi.

 

[Outrageous]

One can also show that (0, 1)^2= -1: (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+1(0))= (-1, 0) which, as above, we identify with the real number -1. We define "i= (0, 1)" so that i^2= -1. And, then, we can write (a, b)= (a, 0)+ (0, b)= (a, 0)+ (b, 0)(0, 1)= a+ bi.

Thank you.
Can I say all the number actually is the combination of real and imaginary number.
As you say" We can "identify" complex numbers of the form (a, 0) with the real number, a"

 

[HallsofIvy]

Yes, that is write. By any definition of "complex" number, we can write any complex number in the form z= a+ bi for some real numbers "a" and "b". That form is also "unique" that is, if z= a<sub>1+ b1i and the same z= a2+ b2i, we must have a1= a2 and b1= b2.

Now, in this problem, attempting to find (2+ 6i)^1/3, they are suggesting that you write the answer as "a+ bi", find (a+ bi)³ and set the "real" and "imaginary" parts equal. For example, (a+ bi)²= (a+ bi)(a+ bi)= a²+ 2abi+ b²i²= (a²- b²+ 3abi. If we wanted to find (1+ i)^1/2, we could now set s²- b²= 1 and ab= 1, giving two equations to solve for a and b.</sub>

 

[Outrageous]

Yes, that is write. By any definition of "complex" number, we can write any complex number in the form z= a+ bi for some real numbers "a" and "b". That form is also "unique" that is, if z= a<sub>1+ b1i and the same z= a2+ b2i, we must have a1= a2 and b1= b2.

Now, in this problem, attempting to find (2+ 6i)^1/3, they are suggesting that you write the answer as "a+ bi", find (a+ bi)³ and set the "real" and "imaginary" parts equal. For example, (a+ bi)²= (a+ bi)(a+ bi)= a²+ 2abi+ b²i²= (a²- b²+ 3abi. If we wanted to find (1+ i)^1/2, we could now set s²- b²= 1 and ab= 1, giving two equations to solve for a and b.</sub>

_{Thanks a lot :!)}