Solving (w - (1+j2))^5 = (32/sqrt(2))(1 + j)

[salman213]

1. Find the solution of the following

(w)^4 = 1w can be a complex number (in polar form)

w^n = re^jntheta (0 <= theta < 2pie)

1 = 1e^j(2pie*k) k = 0, 1, 2 ,3 ...equating the two
----------------------------------------
r = 1

theta*n = 2pie*k
theta = 2pie*k/n for k = 0 theta = 0
for k = 1 theta = 2pie/4
for k = 2 theta = 4pie/4
for k = 3 theta = 6pie/4

so there are 4 roots with magnitude 1 and the angles above.
NOW I am confused on how would I apply a similar approach to a question like the following:

(w - (1+ j2))^5 = (32/sqrt(2))(1 + j)


Any help appreciated!

 

[tiny-tim]

Hi salman213!

Hint: When the RHS had arg = 0, you wrote out the arg in all possible different ways … in that case, arg = 0, 2π, 4π, 6π … and then divided.

Here, the RHS has arg = π/4, so all the possible ways of writing it are … ?

 

[salman213]

hello tim,

so
(w - (1+ j2))^5 = (32/sqrt(2))(1.414)e^j45

w = re^jtheta

1 + j2 = (2.23)e^j63.43so would u say r - 2.23 = 25.82
r= 28.05

i think i just went off track...how to i relate the angle..:S?

 

[tiny-tim]

Hello salman213!

(have a square-root: √ and a pi: π )

(w - (1+ j2))^5 = (32/sqrt(2))(1.414)e^j45

w = re^jtheta

1 + j2 = (2.23)e^j63.43

No, do it one step at a time …

(and where does 2.23 come from? (32)^1/5 = 2)

take the fifth root of each side first …

(w - (1+ j2)) = 2(e^jπ/4)^1/5 = one of five posiible values …

 

[salman213]

but hmmm

so one answer for w = 2(e^jπ/4)^1/5 + (1+ j2) ?

but do i have to some how simplify this so I have w = re^jtheta ? or just leave it ?and for the other answers i think i would just have to add 2pie ? to the angles ? but which angle :( if that is correct?

 

[tiny-tim]

Hello salman213!

but hmmm

so one answer for w = 2(e^jπ/4)^1/5 + (1+ j2) ?

but do i have to some how simplify this so I have w = re^jtheta ? or just leave it ?

You can't just leave it …

it's neither one thing nor the other …

expand the (e^jπ/4)^1/5 as cos + jsin so that the whole thing becomes a + jb

and for the other answers i think i would just have to add 2pie ? to the angles ? but which angle :( if that is correct?

(e^jπ/4)^1/5 has 5 values …

so you add multiples of 2π to … ?

 

[salman213]

to π/4 ?

 

[tiny-tim]

to π/4 ?

Yup!

 

[salman213]

so the solution for w is

w = 2(e^j(2π*k+π/4)^1/5 + (1+ j2) k = 0, 1, 2, 3, and 4 ?


would this be always the case, if you have such solutions to find would u always take the the nth root of the RHS and then add 2pie to the angle?

when would u add for example PIE :S?

 

[tiny-tim]

so the solution for w is

w = 2(e^j(2π*k+π/4)^1/5 + (1+ j2) k = 0, 1, 2, 3, and 4 ?


would this be always the case, if you have such solutions to find would u always take the the nth root of the RHS and then add 2pie to the angle?

when would u add for example PIE :S?

It's clearer if you write it as

w = 2(e^j(π/20 + (2πk/5)) + (1+ j2) k = 0, 1, 2, 3, and 4 …

then you can see exactly where the 2π comes

(and no, it wouldn't be π, it's always 2π )

 

[salman213]

also actually one of the preliminary steps what happened to the √2

in the original question (w - (1+ j2))^5 = (32/sqrt(2))(1 + j)

32/√2 disappeared :(

 

[tiny-tim]

… what happened to the √2

in the original question (w - (1+ j2))^5 = (32/sqrt(2))(1 + j)

32/√2 disappeared :(

Hint: what's e^jπ/4 (in the form a + jb)?

 

[salman213]

ohhkk got ya... it cancels out with the magnitude of that..

so my final answer would be w = 2(e^j(π/20 + (2πk/5)) + (1+ j2)

w = [2cos(π/20 + (2πk/5) + 1] + j[2sin(π/20 + (2πk/5)) + 2] k = 0, 1, 2, 3, and 4 …

so in case i had other powers example in general

(w - complex)^p = complexthe first step is to

w = complex^1/p + complex

and the complex^1/p has an angle which u add 2pie to p timeswould you say that is correct ?

 

[tiny-tim]

so in case i had other powers example in general

(w - complex)^p = complex


the first step is to

w = complex^1/p + complex

and the complex^1/p has an angle which u add 2pie to p times


would you say that is correct ?

Yes!

 

[salman213]

thank you tim!