1. ## complex number simple (wrong thread)

hey guys ive got a problem

If z = 3− i and w = 1+3i, evaluate Im (z∗w)/(2z − 3w) *=conjurate(ithink)

wehn i expand the top line i get 10i/3-11i however it says the answer is 3/13??? ive tired multipling everything by 3+11i however im stuck. any help would be good. thanks

2. Originally Posted by geffman1
hey guys ive got a problem

If z = 3− i and w = 1+3i, evaluate Im (z∗w)/(2z − 3w) *=conjurate(ithink)

wehn i expand the top line i get 10i/3-11i however it says the answer is 3/13??? ive tired multipling everything by 3+11i however im stuck. any help would be good. thanks
$\frac {zw}{2z - 3w} = \frac {(3 - i)(1 + 3i)}{2(3 - i) - 3(1 + 3i)} = \frac {6 + 8i}{3 - 11i}$

now, multiply by $\frac {3 + 11i}{3 + 11i}$

3. If z = 3− i and w = 1+3i, evaluate Im (z∗w)/(2z − 3w) *=conjurate(ithink)
z * w = (3-i)(1+3i) = 3 + 3 + 8i

2z - 3w = 2(3-i) - 3(1+3i) = 6 - 2i - 3 -9i = 3 - 11i

$\frac{6+8i}{3 - 11i} =$

$\frac{(6+8i)(3+11i)}{9+121} = \frac{18+88-42i}{130}\frac{106-42i}{130}$

Question: Does the 'Im' apply to the numerator only or the whole thing??

4. Originally Posted by geffman1
i expand the top line i get 10i/3-11i however it says the answer is 3/13??? ive tired multipling everything by 3+11i however im stuck. any help would be good. thanks
You need to know that $\text{Im} \left( {\overline z w} \right) = 10$ there is no $i$ in that.
$\text{Im}(z)$ is a real valued function.

5. i think it wants you to work out the whole thing, than just write what is the 'imaginary part'. thanks

6. Originally Posted by geffman1
hey guys ive got a problem

If z = 3− i and w = 1+3i, evaluate Im (z∗w)/(2z − 3w) *=conjurate(ithink)

wehn i expand the top line i get 10i/3-11i however it says the answer is 3/13??? ive tired multipling everything by 3+11i however im stuck. any help would be good. thanks
Ever saw "Cool Hand Luke"? "What we have here is a . . . "

That star needs to be on top if you want it to be conjugate like $z^*$ but even that is cheezy. Better to make a bar over it like $\overline{z}$. The $\text{Im}$ and division operators are ambiguous. I would take $\text{Im}$ having a higher precedence than division but that's not what you want. First get the answer in Mathematica. Yea, that's cheezy too but at least it eliminates the ambiguity in notation above. What we then have here is:

$\text{Im}\left(\frac{\overline{z} w}{2z-3w}\right)$

Edit: Make that times two.