
#1
Nov712, 08:33 PM

P: 375

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? 



#2
Nov712, 10:17 PM

P: 66




#3
Nov812, 01:04 AM

P: 375

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




#4
Nov812, 07:02 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890

Complex numberA 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 [itex](0, 1)^2= 1[/itex]: (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 [itex]i^2= 1[/itex]. And, then, we can write (a, b)= (a, 0)+ (0, b)= (a, 0)+ (b, 0)(0, 1)= a+ bi. 



#5
Nov1512, 08:29 PM

P: 375

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" 



#6
Nov1612, 07:48 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890

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[sub]+ b_{1}i and the same z= a_{2}+ b_{2}i, we must have a_{1}= a_{2} and b_{1}= b_{2}.
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)= a^{2}+ 2abi+ b^{2}i^{2}= (a^{2} b^{2}+ 3abi. If we wanted to find (1+ i)^{1/2}, we could now set s^{2} b^{2}= 1 and ab= 1, giving two equations to solve for a and b. 



#7
Nov1612, 08:17 AM

P: 375




Register to reply 
Related Discussions  
taking a number to a complex number power  General Math  3  
Complex number  Precalculus Mathematics Homework  4  
About complex number  Precalculus Mathematics Homework  2  
complex number  Precalculus Mathematics Homework  14  
complex number  General Math  8 