a + jb

[shermaine]

Hi,

How do i solve this equation of (z - 1)^4 = (1 + J)^10 / (1-J) in form of (a + jb)?

 

[dknguyen]

You would solve it like any other equation...isolate z. Then you manipulate the expression to get it into the form of a+jb. One trick is that since j*j = -1, you can multiply expressions like (x+jy) by it's conjugate divided by it's conjugateTHis means you would multiply the expression by (x-jy) / (x-jy) = 1. Since it's equal to one, you aren't changing the expression. YOu are just changing the way it looks.

(x+jy)*(x-jy)/(x-jy) = (x^2+y^2)/(x-jy).

What this lets you do is move the complex j expression between the numerator and denominator. ANd if you do it right, it probably divides itself out, or puts the j in a place where you want it.

 

[Tesla23]

Hi,

How do i solve this equation of (z - 1)^4 = (1 + J)^10 / (1-J) in form of (a + jb)?

Try converting the RHS to polar form and taking the 4th root.

once you have found one solution, try to find the other three.

 

[Miles Prower]

First thing I would do is convert that mess on the right side of the '=' sign to Steinmetz form.

If: C= a + jb (Cartesian) then

R= sqrt(a^2 + b^2)

Theta= arctan(b/a)

C= R /_ Theta (Steinmetz)

That makes exponentiation and division relatively straight forward.

C^x= R^x /_ (Theta * x)

Division:

A= Ra /_ Thetaa

B= Rb /_ Thetab

A / B= (Ra / Rb) /_ (Thetaa - Thetab)

Multiplication:

A X B= (Ra X Rb) /_ (Thetaa + Thetab)

If you want it in Cartesian form:

a= Rcos(Theta)

b= jRsin(Theta)