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Complex Numbers Question- Help Please!

Could someone help me with question 2 please:

w=1+2j

(i) i got: w^2= -1+4j w^3= -9+2j w^4= 15-8j

(ii) i got: p=78/17 and q= 46/17

(iii) i got 1-2j=0

(iv) i have no idea how to do this!

Could someone please check my answers (pic included) as I think my part (ii) might be wrong (the numbers seem very strange, how could I check they are right?) :/

IMG_2084[1].jpg

Many thanks to whoever helps! :smile:

Reply 1

Jooooshy

Part (i) is wrong.. I'm not going to go on from there.

I'll start you off with expanding w^2:

w^2 = (1)(1) +(1)(2j) + (1)(2j) + (2j)(2j)

Write that in the form a + bi, then do question (i) and I'll check the rest :smile:

Reply 2

Davelittle

Original post by Jooooshy

oh god 2*2=4! Silly me :tongue:

Reply 3

Davelittle

Original post by Jooooshy

For (ii) I got p=2 and q=10! :smile:

I don't know how to even approach (iv) though :/

Reply 4

brianeverit

Original post by Davelittle

For (ii) I got p=2 and q=10! :smile:

I don't know how to even approach (iv) though :/

Your p and q are3 correct.
For (iii) Complex roots always occur in conjugate pairs, i.e. if a+bj is a root then so is a-bj
Then for (iv) If you have two roots, a+bj and a-bj then [z-(a+bj)] and [z-(a-bj)] must be factors of the polynomial. So multiplying these together we get a quadratic factor. Divide the polynomial by this quadratic factor to obtain the remaining quadratic factor and solve to find the last two roots. To help you in this part note that if we write the two factors as [z-(a+bj)]=[(z-a)-bj] and [z-(a-bj)]=[(z-a)+bj] we can use the difference of two squares to get (z-a)^2+b^2 for the quadratic factor