From "Linear Algebra Demystified", David McMahon, 2006, problem 2, page 132 and page 235:
$$<\underline{u},\underline{v}> = 2i$$ $$<\underline{u},\underline{w}> = 1 + 9i$$
$$ y = (2\ <3i\underline{u},\ \ \underline{v}>\ ) - <\underline{u},\ \ i \underline{w}> $$
Find value of y:
$$ y = (2(-3i)<\underline{u},\ \ \underline{v}>\ )\ - (i<\underline{u},\ \ \underline{w}>) \\ $$
$$ y = ((-6i)<\underline{u},\ \ \underline{v}>\ )\ - (i<\underline{u},\ \ \underline{w}>) \\ $$
$$ y = ((-6i)(2i) - (i)(1+9i) $$
$$ y = -12i^2 - i-9i^2) $$
$$ y = -21i^2 - i $$
$$ y = 21 - i $$
However, book says the answer is:
$$ -3 -i $$
How did they get that? are they wrong or right? just curious because complex inner product has some strange factoring rules for pulling out constants, being first argument anti-linear, and second argument linear.
If i understand correctly, when pulling the complex constant out of the first argument of dot product then i need to conjugate the imaginary term, if i pull the complex constant out of the second argument of dot product, then i don't need to conjugate it.
and as usual: $i^2 = \sqrt{-1} \sqrt{-1} = -1$
Actually, i'm a little bit confused about this part. The book say that complex vectors have the following inner space properties:
$$ <a\underline{u}, \underline{w}> = a^* <\underline{u}, \underline{w}> $$
$$ <\underline{u}, b\underline{w}> = b <\underline{u}, \underline{w}> $$
And they had a proof in the book that proves it this way...but when i look on the internet, they say its this, instead:
$$ <a\underline{u}, \underline{w}> = a <\underline{u}, \underline{w}> $$
$$ <\underline{u}, b\underline{w}> = b^* <\underline{u}, \underline{w}> $$
(my opinion, is that mathworld wrote it wrong then wikipedia copied it, and then everybody else believed them)
$$2<3iu,v>-<u,iw>=-6i\cdot2i-i(1+9i)=12-i+9=21-i,$$ which says that the book answer is wrong and you are right.