Operations on complex numbers (1 Viewer)

[YBK]

Hey, I'm having troubles with this question:
(question 10 from exercise 2.2A in terry lee)


Simplify the following by expressing in the form a + ib, where a, b are real

b) 1/(1 + cosx + isinx)

I get up to this step:

1+cosx - isinx / (2+2cosx)

but can't simplify it further


Thanks

 

[Riviet]

Hey, I'm having troubles with this question:
(question 10 from exercise 2.2A in terry lee)


Simplify the following by expressing in the form a + ib, where a, b are real

b) 1/(1 + cosx + isinx)

I get up to this step:

1+cosx - isinx / (2+2cosx)

but can't simplify it further


Thanks

Hmm... you must have an older edition because i have the newest (6th edition) and it has your question in ex 2.2 of Q13 c) instead.

1/(1+cosx+isinx) = (1+cosx-isinx)/[(1+cosx)²+sin²x]=

1+cosx-isinx=
2+2cosx

1+cosx-isinx=
2(1+cosx)

1+cosx ___ isinx (yes that's a large minus sign lol)
2(1+cosx) 2(1+cosx)

=1/2 - i/2 . (sinx/[1+cosx])

=1/2 - i/2 . [2sin(x/2)cos(x/2)]/[2cos²(x/2)]

=1/2 . (1 - itan[x/2])

There are a few sneaky tricks, so ask if you dont understand anything.

 

[YBK]

=1/2 - i/2 . [2sin(x/2)cos(x/2)]/[2cos2(x/2)]




I don't understand that step...

is it some identity that I haven't learnt yet or something????

 

[webby234]

He's just used the identity
sin 2x = 2sinxcosx

 

[YBK]

He's just used the identity
sin 2x = 2sinxcosx

cool, how about the denominator?

 

[YBK]

btw, Riviet, I have a 5th edition.. got it for a good price

 

[icycloud]

cool, how about the denominator?

cos(2x) = 2cos²(x) - 1
Thus, cos(x) = 2cos²(x/2) - 1

So, cos(x) + 1 = 2cos²(x/2)

 

[icycloud]

btw, Riviet, I have a 5th edition.. got it for a good price

I wanted the 5th edition! The typesetting is so much better than the 6th edition.

 

[YBK]

cos(2x) = 2cos²(x) - 1
Thus, cos(x) = 2cos²(x/2) - 1

So, cos(x) + 1 = 2cos²(x/2)

yay!

*now tries to figure out how the last line got there by checking identities in excel book*


btw, how old is the 5th edition? i hope it's not out of date or something...

 

[YBK]

wooo!

i get it

it's because the 2cos (x/2) cancels with the 2cos^2(x/2) leaving a sin (x/2) / cos (x/2) which is tan (x/2)

 

[Riviet]

Nice work guys.

 

[Trebla]

Hey, I'm having troubles with this question:
(question 10 from exercise 2.2A in terry lee)


Simplify the following by expressing in the form a + ib, where a, b are real

b) 1/(1 + cosx + isinx)

I get up to this step:

1+cosx - isinx / (2+2cosx)

but can't simplify it further


Thanks

You could also use t-forumlae (if you remember them) I think.....

 

[KeypadSDM]

The t-formulae make the question very simple [to an extent], but the manipulative algebra employed using the half angle rules is much more important.

An easier method would be to note that when you've got a 1 on the bottom, it's nice to remove it using the half angle rules, and using the knowledge that most questions come out very simply:

1/(1 + cosx + isinx)
= (2cos²[x/2] + 2isin[x/2]cos[x/2])^-1
=1/(2cos[x/2]) * 1/(cos[x/2] + isin[x/2])
[Note that the right hand multiplier is an inverse of a unit complex number]
=(cos[x/2] - isin[x/2])/(2cos[x/2])
=1/2 * (1 - itan[x/2])

 

[YBK]

Cool, thanks everyone...!

I'll try to understand KeyPad's way

 

[insert-username]

btw, how old is the 5th edition? i hope it's not out of date or something...

It's a maths book. They never go out of date.


I_F

 

[Riviet]

It's a maths book. They never go out of date.

Lol that is true, the 5th edition would not be that old probably around 2000-04 or so

 

[buchanan]

It's a maths book. They never go out of date.

Yes they do. I solved the problem in 1 line using hyperbolic tangent:

I bet you $50 that wasn't in your crummy out-of-date book!