Spec '10 - Help forum

[TrueTears]

(1+i)^(1/2) = 1 + 1/2(i) -1/8 i^2 + 1/16i^3 -5/128i^4  +7/256i^5 - 21/1024i^6 +33/2048i^7... = 1+1/2(i) +1/8 -1/16(i) -5/128 + 7/256(i)+ 21/1024 - 33/2048(i)...

the non i terms form a series with sum rt[2(rt[2]+1)]/2

the terms with i form a series with sum rt[2(rt[2]-1)]/2(i)

Since both are converging series, their sum is also converging rt[2(rt[2]+1)]/2+rt[2(rt[2]-1)]/2(i) = (1+i)^(1/2)

[Mao]

(1+i)^(1/2) = 1 + 1/2(i) -1/8 i^2 + 1/16i^3 -5/128i^4  +7/256i^5 - 21/1024i^6 +33/2048i^7... = 1+1/2(i) +1/8 -1/16(i) -5/128 + 7/256(i)+ 21/1024 - 33/2048(i)...

the non i terms form a series with sum rt[2(rt[2]+1)]/2

the terms with i form a series with sum rt[2(rt[2]-1)]/2(i)

Since both are converging series, their sum is also converging rt[2(rt[2]+1)]/2+rt[2(rt[2]-1)]/2(i) = (1+i)^(1/2)

A part of me wants to slap you. This has been in no way helpful to students.

[/0]

it is just extended to C

maybe u can enlighten me with the proof, but i am happy to just apply it for now, will look at the proof once i know how to utilise it.

spoken like a true physicist 


(at least that's what some people in /math/ keep telling me)

[TrueTears]

doesn't hurt to learn something new these days

[Chavi]

How would i figure out the intersection of two circles?
4. Find the coordinates of the point of intersection of the following loci: and


Thanks

[brightsky]

(1)

(2)

From (1),



(3)

From (2),









(4)

Substitute (4) into (3):





When















Using quadratic formula,

or

Do the same for when

[the.watchman]

Wouldn't it have been a whole lot easier to just take eqns 2 and 3 and let them equal one another

Then you would have gotten , you can sub this into the originals then.

Or have I made a mistake?

[brightsky]

Wouldn't it have been a whole lot easier to just take eqns 2 and 3 and let them equal one another

Then you would have gotten , you can sub this into the originals then.

Or have I made a mistake?

Yeah, I think you have. That would be incorrect logic as by doing that you've solved for both equations, and substituting that into both equations, you would get a weird result like . Haven't tried it yet though, so I may be wrong..

[Mao]

Wouldn't it have been a whole lot easier to just take eqns 2 and 3 and let them equal one another

Then you would have gotten , you can sub this into the originals then.

Or have I made a mistake?

Yeah, I think you have. That would be incorrect logic as by doing that you've solved for both equations, and substituting that into both equations, you would get a weird result like . Haven't tried it yet though, so I may be wrong..

In this case, it works.

A trivial result would be obtained if a substitution is made in the equation it was derived from. However, this substitution is not derived independently from either equations, hence the substitution is valid.

However, you are correct in in being careful. If not done right, this method can give trivial results.

[the.watchman]

Whew! Lucky again!

[Chavi]

Can I please get some help with this question - according to the answer, the function randomly stops between [-1, 1]. Can somebody please explain? Thanks.

16a. I denoted the complex conjugate of the number ,
Sketch on an argand diagram:

[fady_22]

The function "stops" because in these values of re(z), the Arg(z)(with capital A, not lower case as you said it was-- this makes a big difference) is greater than pi/2.

[luken93]

hello all, in yr 11 and just wondering if you could give us a hand with this q.
i know it will be easy for all you geniuses but i'm having trouble with them.

Solve each of the following pairs of simultaneous equations for x and y
(a + b)x + cy = bc
(b + c)y + ax = -ab

Thanks, and btw I'm not that great on latex hence the normal type.

[m@tty]

Those two would've just worked if you placed "tex" and "/tex", each enclosed in square brackets i.e. [...].



From [1] 


Sub [3] into [2]





Sub that in to [2]




You should always check the solutions you found, by subbing both in to an equation.



Therefore your solutions are correct.

These questions are just a pain.

[TrueTears]

What you could do is, you can set up a matrix and use guass-jordon elimination =)