Help MX2 bebz x2 (1 Viewer)

Dylanamali

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a) Let z = a+ib be a complex number, where a and b are real numbers. Prove that the reciprocal of z (i.e. 1/z) is a complex number.

z = a+ib
1/z = 1/a+ib = 1/a+ib x a-ib/a-ib = a-ib/a^2+b^2
= a/a^2+b^2 + i b/a^2+b^2

and therefore 1/z is complex as it has a complex component Im (z) = b/a^2+b^2.

Is that the correct way to prove it?


b) Express 3^i in the form a+ib where a and b are real numbers.


c) Find the value of the constants a and b to ensure the following function is continuous for all real values of x:

f(x) = x^2-4/x-2 for x<2
= ax^2-bx+3 for 2≤x<3
= 2x-a+b for x≥3



THANKS BABES AND DEF REPS FOR ALL!
 

Carrotsticks

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a) Yes. Provided Z is not a purely real number and not equal to 0, its reciprocal 1/Z will be complex. You can think about this in terms of De Moivre's Theorem. By finding the reciprocal, you are raising it to the power of -1, which is the same as multiplying the angle by -1. If the angle is not equal to 0 or pi (ie: Z is purely real), then the reciprocal will be complex.

If you really want to be specific, we could always argue that no matter what transformation we perform upon Z, the output will be a complex number (no matter whether it's purely real or not) because the field of Real numbers is a subfield of Complex numbers.

b)



c)

For the function to be continuous, we must prove that the left-limit is equal to the right-limit at the points x=2 and x=3 (because this is where all the funky stuff happens). We must also show that there exists an output when x=2 and 3 (or else the limit could exist, but there's an open circle there).

So for the x E [2,3) part:



Well really, since the inequalities aren't strict, we need not prove the existence of outputs at x=2 and x=3.
 

Dylanamali

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Thank you carrot, me loves you very much :)
unfortunately can't rep you coz I have to share it around =/

Everything makes perfect sense.
Only difficulty i'm having is understanding c)

Unfortunately because I don't know how to use latex you misinterpreted the question. The first part is:

f(x) = (x^2-4)/(x-2) for x<2

Following that through your working, you will end up with 4a-2b+3 = DNE. Where do you go from there?

Also in regards to your last statement, could you clarify that further?
I understand that for a function to be continuous f(c) must exist, lim f(x) must exist and f(c) must = lim f(x), just not sure how to go about using that to solve this problem. I understand how you went about your steps, but why isn't there a usage of f(c).

Thank you so much, really appreciate it :)
 

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