Solving a Complex Vector Problem: Finding the Magnitude of ix+y

[fredrick08]

Homework Statement

let x=(i,1,1) and y=(1,i,2)

find the magnitude of ix+y

The Attempt at a Solution

i(i,1,1)+(1,i,2)=(0,2i,2+i)

therefore mag sqrd = 0^2+(2i+2(-i))+(2+i)(2-i)=9 => magnitude = 3?

is this rite? i have completely forgotten when to take the conjugates... idk if I am not meant to take the conjugate of 2i, but i have to for 2+i... please can anyone reassure me.

 

[Office_Shredder]

It doesn't look like you did it right... why don't you start by writing down the definition of the magnitude of a vector? Then remember that $|a|^2 = a \bar{a}$

 

[fredrick08]

magnitude of a vector is its length, normally would do pythagouras, but for complex the magnitude is sqr root of the vector*conjugate. i think

 

[Office_Shredder]

So why don't you write that out explicitly? That's not what you did in your original post. Work it out slowly

 

[fredrick08]

ok well i(i,1,1)dotproduct(1,i,2)=(-1,i,i)dotproduct(1,i,2)=(-1+1)i+(i+i)j+(2+i)k=(0,2i,2+i)

sqr root of that taking conjugates..(which not not sure about) = root(0^2+(2i*-2i)+((2+i)(2-i)))=root(0+4+5)=root(9)=3? i can't take it any slower then that, please can u tell me where i am going wrong

 

[fredrick08]

or r u sayingi have to take the conjugate when i multiply x by i?

then it would be root((1,i,i)dot(1,-i,-i)+(1,i,2)dot(1,-i,2))=root((1+1+1)+(1+1+4))=root(9)=3?? omg i don't know

 

[fredrick08]

anyone have any ideaS?

 

[fredrick08]

please anyone?

 

[fredrick08]

no one?

 

[Mark44]

or r u sayingi have to take the conjugate when i multiply x by i?

I don't believe so. Your ix + y is the vector you want the magnitude of, so it's the one you want to multiply by its complex conjugate.

then it would be root((1,i,i)dot(1,-i,-i)+(1,i,2)dot(1,-i,2))=root((1+1+1)+(1+1+4))=root(9)=3?? omg i don't know

I also get 3 for the magnitude of ix + y.

 

[fredrick08]

ok thankyou very much = ) yes i thought so, just havnt done this stuff in so long, lost confidence.