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A little linear algebra and complex analysis

Hi, I have two problems that I cant seem to get my head around. Any help would be appreciated!

The first is linear algebra: Let X be the set of all infinite sequences (a_0,a_1,...,a_n,...) of real numbers a_i which satisfy a_(i+2)=a_(i+1)+a_i. I've shown that X is a vector space but I also have to find a basis of X. How would I go about that?

The other problem is: Sketch the set {z is a complex number : 1<|2z-6|<2}. This has confused the hell out of me. I've tried substituting in z=a+bi and then getting 1<4a^2-24a+36+2b^2<2, is this the right thing to do? If so, where do I go from here?

Thanks!
Original post by mthpossom
Hi, I have two problems that I cant seem to get my head around. Any help would be appreciated!

The first is linear algebra: Let X be the set of all infinite sequences (a_0,a_1,...,a_n,...) of real numbers a_i which satisfy a_(i+2)=a_(i+1)+a_i. I've shown that X is a vector space but I also have to find a basis of X. How would I go about that?


Looking at the definition of the sequence, how many terms are required to uniquely determine the sequence, and go from there.


The other problem is: Sketch the set {z is a complex number : 1<|2z-6|<2}. This has confused the hell out of me. I've tried substituting in z=a+bi and then getting 1<4a^2-24a+36+2b^2<2, is this the right thing to do? If so, where do I go from here?

Thanks!


Don't know how you worked out that expression, though I'd guess you were attempting to square the modulus. It wouldn't be 2b^2, and the limits of the inequality would change as well.

However, going back to your original inequality, I'd start by dividing through by 2.

Then if you can't see the solution, consider the simpler case.

a < |z| < b

What's that set?

Drawing it is easier than doing it algebraically.
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Avatar for Hasufel
Hasufel
16
the region, for the second question is an annular region (not including the boundary), it`s the same as (dividing inequality by 2):

1/2<|z-3|<1
(edited 11 years ago)

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