Maths Help !

3fullback

Any Genius want to help me.

Doing a few past papers but got extremely stuck on this one, if anyone has any ideas or could do with the practice themselves, I'd love if someone could show me how to do it. Thanks Guys.

Let u =3 - 6i


(B)(ii) Show .


Ps. Its the 1999 OL Paper1 Q4. b) ii)

boblong

I'm presuming you're dealing with imaginary numbers and i isn't just another variable.
If someone wants to Latex it up fine, but I'm in a bit of a rush!

i(3-6i)+(3-6i)/i
3i-6i^2+3/i-6
3i+6+3/i-6
3i+3/i
(3i^2+3)/i
(-3+3)/i
0/i
=0

Could be wrong!

irish_man

i^2 (i squared) is -1 so you just multiply and you get
-u+u=0
u=u

3fullback

boblong wrote: »

I'm presuming you're dealing with imaginary numbers and i isn't just another variable.
If someone wants to Latex it up fine, but I'm in a bit of a rush!

i(3-6i)+(3-6i)/i
3i-6i^2+3/i-6 <

---

3i+6+3/i-6
3i+3/i
(3i^2+3)/i
(-3+3)/i
0/i
=0

Could be wrong!

Where did 2+3 come from ??

Would you not cross multiple 3-6i x i
.........................................i......i
And get i^2+3i-6i^2 (i^2 = -1)
-1+3i+6

So now bring in the first part of the eqn

3i-6i-1+3i+6
6i+5=0

Now is when i don't know what else to do, i don't even know if i'm doing it the right way ?

Anyone any ideas?

jumpguy

3fullback wrote: »

Any Genius want to help me.

Doing a few past papers but got extremely stuck on this one, if anyone has any ideas or could do with the practice themselves, I'd love if someone could show me how to do it. Thanks Guys.

Let u =3 - 6i


(B)(ii) Show .


Ps. Its the 1999 OL Paper1 Q4. b) ii)

ui + u/i = 0
i(3-6i) + 3-6i/i = 0
3i + 6 + 3/i - 6 = 0 (Multiply everything now by i, remember i^2 = -1)
-3 + 6i + 3 - 6i = 0
-3+3 +6i-6i = 0
0=0

Any questions, and I'm happy to answer them.

boblong

That's ^ pretty much what I wrote. I think you're misreading it OP, it's 6i^2+3/i not 2+3

irish_man

mines right up there ^

3fullback

jumpguy wrote: »

ui + u/i = 0
i(3-6i) + 3-6i/i = 0 <

---

3i + 6 + 3/i - 6 = 0 (Multiply everything now by i, remember i^2 = -1)
-3 + 6i + 3 - 6i = 0
-3+3 +6i-6i = 0
0=0

Any questions, and I'm happy to answer them.

- by a - not a plus ???????

-6i^2

-6(-1)

+6 ????????

3fullback

boblong wrote: »

That's ^ pretty much what I wrote. I think you're misreading it OP, it's 6i^2+3/i not 2+3

sorry mate misread that bit alright, still though its not working out

boblong

They are haha, both solutions posted work. Is your confusion based in imaginary numbers?

jumpguy

3fullback wrote: »

- by a - not a plus ???????

-6i^2

-6(-1)

+6 ????????

Since you're pointing to my second line, I'll explain it.
ui + u/i = 0 (1)
i(3-6i) + 3-6i/i = 0 <

---

(2)
3i + 6 + 3/i - 6 = 0 (Multiply everything now by i, remember i^2 = -1) (3)

The first line (1) all I did was merely write down the question, obviously.

In the second line, I plugged in the value for u.

In the third line, I multiplied out the brackets (that was i(3-6i) ). Also in the third line, where I had 3-6i/i (that is, 3 minus 6i ALL over i), I divided the i into -6i and got -6. Because I can't divide i into 3, I had to leave 3 all over i.

In the 4th line, I multiplied everything by i.


I'm not sure where you're getting your question from, but I hope my further elaboration explains things.

NOTE: The way I've answered the question is slightly different from the way boblong did it.

3fullback

boblong wrote: »

I'm presuming you're dealing with imaginary numbers and i isn't just another variable.
If someone wants to Latex it up fine, but I'm in a bit of a rush!

i(3-6i)+(3-6i)/i
3i-6i^2+3/i-6 ......." this should be a +6 "......
3i+6+3/i-6
3i+3/i
(3i^2+3)/i
(-3+3)/i
0/i
=0

Could be wrong!

edit on your quote

that would then leave you with +6+6 = 12

3fullback

jumpguy wrote: »

Since you're pointing to my second line, I'll explain it.
ui + u/i = 0 (1)
i(3-6i) + 3-6i/i = 0 <

---

(2)
3i + 6 + 3/i - 6 = 0 (Multiply everything now by i, remember i^2 = -1) (3)

The first line (1) all I did was merely write down the question, obviously.

In the second line, I plugged in the value for u.

In the third line, I multiplied out the brackets (that was i(3-6i) ). Also in the third line, where I had 3-6i/i (that is, 3 minus 6i ALL over i), I divided the i into -6i and got -6. Because I can't divide i into 3, I had to leave 3 all over i.

In the 4th line, I multiplied everything by i.


I'm not sure where you're getting your question from, but I hope my further elaboration explains things.

NOTE: The way I've answered the question is slightly different from the way boblong did it.

but with complex numbers for something like dividing by i you are meant to cross multiply , thats what i thought

boblong

The "i"s cancel. I wrote it up properly for you http://docs.google.com/View?id=dg6h9bvw_152dsjj6ngd

jumpguy

3fullback wrote: »

but with complex numbers for something like dividing by i you are meant to cross multiply , thats what i thought

There's an easier way to do this that might be less confusing. You don't even have to plug in the value for u.

ui + u/i = 0

Multiple everything by i

ui^2 + u = 0

Remember i^2 = -1

-u + u = 0

0=0

3fullback

Guys thanks so much for the help.
boblong that google doc really helped , way easier to uderstand.

lets hope the L.C. goes a bit better:p

Thanks guys