# Inner Product

## Homework Statement

Let v,w be vectors in a complex inner product space such that ||v|| = 1,
||w|| = 3 and <v,w> = 1 + 2i. Find ||v + iw||.

## Homework Equations

The properties of an inner product.

## The Attempt at a Solution

I figured
$$||v+iw||^2$$ = <v+iw,v+iw>

Then using the properties of the inner product, I broke it up;

$$||v+iw||^2$$ = <v,v+iw>+<iw,v+iw>

and <v,v+iw> = <v,v> + <v,iw> and using the 'conjugate symmetry'
= <v,v> - i<v,w>
= 1-i(1+2i)
= 3-i

Now for <iw,v+iw>;
=i<w,v+iw>
=i{<w,v>+<w,iw>}
=i{1-2i - i<w,w>} (since <v,w>=$$\overline{<w,v>}$$)
=i+2-9i
=2-8i

Now combining it all i get $$||v+iw||^2$$ = 5-9i

But this is supposed to be the square of the magnitude of v+iw.. so it should be a real number right?

Can somebody help point out where I have gone wrong?
Cheers.

HallsofIvy
Homework Helper

## Homework Statement

Let v,w be vectors in a complex inner product space such that ||v|| = 1,
||w|| = 3 and <v,w> = 1 + 2i. Find ||v + iw||.

## Homework Equations

The properties of an inner product.

## The Attempt at a Solution

I figured
$$||v+iw||^2$$ = <v+iw,v+iw>

Then using the properties of the inner product, I broke it up;

$$||v+iw||^2$$ = <v,v+iw>+<iw,v+iw>

and <v,v+iw> = <v,v> + <v,iw> and using the 'conjugate symmetry'
= <v,v> - i<v,w>
= 1-i(1+2i)
= 3-i

Now for <iw,v+iw>;
=i<w,v+iw>
=i{<w,v>+<w,iw>}
=i{1-2i - i<w,w>} (since <v,w>=$$\overline{<w,v>}$$)
=i+2-9i
Here is your error. i(1- 2i- 9i)= i+ 2+ 9. You forgot the second i on the last term.

=2-8i

Now combining it all i get $$||v+iw||^2$$ = 5-9i

But this is supposed to be the square of the magnitude of v+iw.. so it should be a real number right?

Can somebody help point out where I have gone wrong?
Cheers.

## The Attempt at a Solution

oops, thanks a lot halls!