Find <x,y> in Inner Product Spaces: 8+5i

snesnerd
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Let x = (2,1+i,i) and y = (2-i,2,1+2i). Find <x,y>

So my work is the following:

2(2-i) + (1+i)2 + i(1+2i) = 4+i, but my book says the correct answer is 8+5i. Hmmm what am I doing wrong?
 
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snesnerd said:
Let x = (2,1+i,i) and y = (2-i,2,1+2i). Find <x,y>

So my work is the following:

2(2-i) + (1+i)2 + i(1+2i) = 4+i, but my book says the correct answer is 8+5i. Hmmm what am I doing wrong?

The inner product in complex spaces involves taking the complex conjugate of one of the vectors before you do the inner product. Look up your definition of inner product. Otherwise it's not a real inner product. <x,x> won't be greater than or equal to 0.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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