Ü
Sorry I’m A retardcel
Use L hopital
Can only use if 0/0 or infinity/infinity bhai
nigga just use photomath solve your own fuckin problems 
Some problems can't be solved there thoughnigga just use photomath solve your own fuckin problems
Alright, let me work on this. Give me a second.
The first thing we do is isolate z squared. Because we have z squared divided by z squared we can effectively remove z squared entirely, from what I know of my highschool level algebra.
We can't solve the parenthesis yet, so we just ignore them for now. 2 * iz divided by z - 3i requires us to divide by 3 and divide by 2 in order to solve it on both sides. that gives us 2/iz and z -3/i. Problem now looks like this:
(1 - 3i)z i/-3
----------------
3 - iz/2
However, to divide one side by something we have to divide the other side by two as well. So, it actually looks more like this:
(1 - 3i)z i/-3/2
----------------
3 - iz/2/3
2 divided by 3 is 0.6 everlasting, while 3 divided by 2 is 1.5. So we now have:
(1 - 3i)z i/1.5
----------------
3 - iz/0.6~
Now we want to multiply both sides by that, cancelling them out. I might be messing up here? Maybe we were supposed to multiply first instead. It's been ages since I did algebra, god
anyway
Point is, we now lose that and just have iz and z i. Now we want to divide both sides by i!
This gives us:
(1 - 3i)z
----------------
3
Reason why we want to do that is the i gets in the way of the equation, so we want to get rid of it completely. Once we've done that, we can probably do the parenthesis? So we divide both sides by that, giving us
iz
----------------
3/(1 - 3)
So we divide two by three, giving us 0.6 everlasting divided by iz. I hope I did that right, would absolutely reccomend getting someone to doublecheck, I have not done algebra in forever.
We can't solve the parenthesis yet, so we just ignore them for now. 2 * iz divided by z - 3i requires us to divide by 3 and divide by 2 in order to solve it on both sides. that gives us 2/iz and z -3/i. Problem now looks like this:
(1 - 3i)z i/-3
----------------
3 - iz/2
However, to divide one side by something we have to divide the other side by two as well. So, it actually looks more like this:
(1 - 3i)z i/-3/2
----------------
3 - iz/2/3
2 divided by 3 is 0.6 everlasting, while 3 divided by 2 is 1.5. So we now have:
(1 - 3i)z i/1.5
----------------
3 - iz/0.6~
Now we want to multiply both sides by that, cancelling them out. I might be messing up here? Maybe we were supposed to multiply first instead. It's been ages since I did algebra, god
anyway
Point is, we now lose that and just have iz and z i. Now we want to divide both sides by i!
This gives us:
(1 - 3i)z
----------------
3
Reason why we want to do that is the i gets in the way of the equation, so we want to get rid of it completely. Once we've done that, we can probably do the parenthesis? So we divide both sides by that, giving us
iz
----------------
3/(1 - 3)
So we divide two by three, giving us 0.6 everlasting divided by iz. I hope I did that right, would absolutely reccomend getting someone to doublecheck, I have not done algebra in forever.
Guess it's wrong bhai, answer is 3 - i/4
Huh. I still don't get it
Guess it's college level and I've only done highschool level tho
Common root is 3i, therefore you can divide the polynomial and write it as (x - 3i)(x + i) and take the (x - 3i) out, since it's a common factor.Huh. I still don't get it
Guess it's college level and I've only done highschool level tho
Therefore x + 1/x + i
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