Question

Complex numbers z1=(3a+2) + j(3b-1) and z2=(b+1)-j(a+2-b). Write down the real and imaginary parts of z1 and z2. I understand the Real numbers from the complex numbers, but how do you obtain the fraction number of Re(z1) = 3a+2=5/7? and Re(z2) = b+1=5/7? i.e. where does the 5/7 come from?

Answer 1

what do the 'j' stand for?

Answer 2

are your j, the i?

Answer 3

sorry yes, in engineering we use j instead of i.

Answer 4

ah ok, like vectors?

Answer 5

Yeah just to confuse things!

Answer 6

that's so odd... yuck

Answer 7

anywho, so the only info is this: \[z_1=(3a+2) + i(3b-1)\]\[z_2=(b+1)-j(a+2-b)\]

Answer 8

correct?

Answer 9

ignore that j

Answer 10

and you are told to identify the \(Re(z_1)\) that's it?

Answer 11

Yes. And for the first part of the question I had to find the real numbers of a and b; which were a=-3/7 & b=-2/7. but I cant work out the second part of the question which asks me to find the Real and Imaginary parts of z1 and z2.

Answer 12

plug and chug?

Answer 13

Tried but didnt work.

Answer 14

ok let's check z_1 first

Answer 15

so 3a+2=3(-3/7)+14/7

Answer 16

-9+14=5 so 5/7

Answer 17

follow?

Answer 18

No where did you get 3z+2=3(-3/7)+14/7. Where did the 14/7 come from?

Answer 19

14/7=2

Answer 20

the re(z_1) is any part without an i

Answer 21

Oh der to me! Thanks so much that clears it up. I should be ok from there for the Imaginary part. Thanks again!

Answer 22

np just remember to use the LCD and you'll be fine

Answer 23

LCD?

Answer 24

the least common denominator for fractions ie. ⅔+½=4/6+3/6 =7/6

Answer 25

Yes the simple things are throwing me! thanks heaps!

Answer 26

haha they always do, for me it's 7*8 and 6*8, I can't ever remember them