Complex number question (1 Viewer)

[YBK]

Hey, this question is reallly annoying me... no idea :S We've just started complex numbers...

Prove the following results about complex conjugates:

c) complex conjugate z1 z2 = complex conjugate z1 . complex conjugate z2

That is question 3 c in the cambridge 4u book

do you replace z1 by (a+ib) and z2 by (c+id) ???

thanks for any help

 

[richz]

ok

u prove LHS=RHS

try do this let z₁=a+ib
and let z₂=c+id

and u know conjugates are z₁(bar)=a-ib
z₂(bar)=c-id

 

[YBK]

ok

u prove LHS=RHS

try do this let z₁=a+ib
and let z₂=c+id

and u know conjugates are z₁(bar)=a-ib
z₂(bar)=c-id

wow, is that all?

thanks

 

[sikeveo]

just make up values for z1 and z2, find the corresponding conjugates, then manipulate both so that you can go rhs = lhs and state that the identity is true.

 

[pLuvia]

Here's what I got, since I use Cambridge as well, oh and you mean Q4 (c)

Let z₁ = a+bi
Let z₂ = x+yi

z₁z₂
= (a+bi)(x+yi)
= (ax + ayi + bxi + byi²
= (ax-by) + (ay+bx)i

(z₁z₂)bar
= (ax-by) - (ay+bx)i

(z₁)bar . (z₂)bar
= (a-bi)(x-yi)
= ax-ayi-bxi+byi²
= (ax-by) - (ay+bx)i

.: (z₁z₂)bar = (z₁)bar . (z₂)bar

 

[YBK]

Here's what I got, since I use Cambridge as well, oh and you mean Q4 (c)

Let z₁ = a+bi
Let z₂ = x+yi

z₁z₂
= (a+bi)(x+yi)
= (ax + ayi + bxi + byi²
= (ax-by) + (ay+bx)i

(z₁z₂)bar
= (ax-by) - (ay+bx)i

(z₁)bar . (z₂)bar
= (a-bi)(x-yi)
= ax-ayi-bxi+byi²
= (ax-by) - (ay+bx)i

.: (z₁z₂)bar = (z₁)bar . (z₂)bar

oops.. yeah i did mean question 4 (c) question 3, i could do surprisingly

cool, thanks dude
but I thought we had to prove LHS = RHS with manipulation of only one side... but yeah, I kinda get it... i will try proving LHS=RHS before I sleep, and if I fail again will ask teacher tomorrow... I can't do any of question 4!!! except of part a) b)

asked teacher... she said it can only be done your way

 

[c0okies]

xD Thank you for the solution!!!