Mathematicians needed

[lostclause]

EDIT: Solved thanks to Lukeje and waxwingslain.

I'm not sure if this is the right sort of thing to discuss (if it isn't, mods please lock this) but I can't exactly google it so I'm hoping the escapist community can help.

Warning: Complex maths ahead, I'm hoping someone here can help me with this because I must be wrong.

I'm currently studying for some school exams and one of the topics is complex numbers. One of the practice questions read like this:
z= i / ( 1 - i ) to be converted to the from a+bi where a and b are real. ( i = square root of -1 of course)

I've given my solution and the way that was given in the answers. ^ means 'to the power of' in this case.

Spoiler: My way of solving it

z= ( [ 1 - i ] / i ) ^ -1
= ( [ 1 / i ] - [ i / i ] ) ^ -1
= ( [ 1 / i ] - 1 ) ^ -1
= ( i / 1 ) - ( 1 / 1 )
= i - 1

Spoiler: Given way of solving it

z= [ i / ( 1 - i ) ] x [ ( 1 + i ) / ( 1 + i ) ]
= i ( 1 + i ) / ( 1 - i^2 )
= i ( 1 + i ) / ( 1 - -1 )
= i ( 1 + i ) / 2
= ( i + i^2 ) / 2
= ( i - 1 ) / 2

Since both seem to be correct you end up with the conclusion 1=2. Now I'm sure there must be some rule I'm breaking but I don't know what it is.

Thanks for any help you can give and sorry if the formatting isn't very good, this forum is geared for maths equations. Something for the suggestion thread perhaps

 

[Iseryn]

How did you get from the third line to the fourth line in your version?

 

[waxwingslain]

i ^-1 = -i. Your first step would probably be fine for real numbers but not for imaginaries.

 

[lostclause]

I've seen this before. All I remember that it somewhere zero is divied or something, giving you 1=2. If I find the equation that was given with my point I'll post it.

I've seen that too, it's entirely different from this. This is something I stumbled on accidently when practising. If you look, there is no place where I divide by zero. Like that one I'm sure this is flawed in some way but I don't know how, that's what I'm trying to find out.

 

[Lukeje]

You've misused reciprocals;

( [ 1 / i ] - 1 )[sup]-1[/sup] =/= ( i / 1 ) - ( 1 / 1 )

 

[lostclause]

How did you get from the third line to the fourth line in your version?

Put everything to the power of -1 (reciprical)

i ^-1 = -i. Your first step would probably be fine for real numbers but not for imaginaries.

But does that matter since I raise it to the power of -1 twice (which would bring me to the same point)?

 

[Iseryn]

How did you get from the third line to the fourth line in your version?

Put everything to the power of -1 (reciprical)

Like Lukeje said, reciprocals don't work like that.

 

[The_root_of_all_evil]

Also i/i1 in the same way that 0/01

 

[HyenaThePirate]

You've misused reciprocals;

( [ 1 / i ] - 1 )[sup]-1[/sup] =/= ( i / 1 ) - ( 1 / 1 )

This.
Monitor your reciprocals. There really is no shortcut for this, so just get used to working it out the long way, every time... besides it's good practice for higher level maths later on or higher level sciences such as Physics and Chemistry which are very math intensive.

 

[waxwingslain]

How did you get from the third line to the fourth line in your version?

Put everything to the power of -1 (reciprical)

i ^-1 = -i. Your first step would probably be fine for real numbers but not for imaginaries.

But does that matter since I raise it to the power of -1 twice (which would bring me to the same point)?

Yes. The point is that, raising a complex number to the power of -1 does more than just flip the fraction, so your first equality isn't correct.

 

[lostclause]

Yes. The point is that, raising a complex number to the power of -1 does more than just flip the fraction, so your first equality isn't correct.

So it should be this?
z= ( -i [ 1 - i ] ) ^ -1
= ( 1 - i ) ^ -1
= i - 1

 

[lostclause]

You've misused reciprocals;

( [ 1 / i ] - 1 )[sup]-1[/sup] =/= ( i / 1 ) - ( 1 / 1 )

What does it equal? Even with waxwing's correction it gets to i - 1
Sorry if I seem to be slow on the uptake, I'm new to this.

 

[Lukeje]

You've misused reciprocals;

( [ 1 / i ] - 1 )[sup]-1[/sup] =/= ( i / 1 ) - ( 1 / 1 )

What does it equal? Even with waxwing's correction it gets to i - 1
Sorry if I seem to be slow on the uptake, I'm new to this.

(1 / i) - 1 = i / (1 - i).
Thus that method of tackling the problem doesn't get you anywhere.

 

[lostclause]

You've misused reciprocals;

( [ 1 / i ] - 1 )[sup]-1[/sup] =/= ( i / 1 ) - ( 1 / 1 )

What does it equal? Even with waxwing's correction it gets to i - 1
Sorry if I seem to be slow on the uptake, I'm new to this.

[1 / i] - 1 = i / (1 - i).
Thus that method of tackling the problem doesn't get you anywhere.

I see. Thanks.

 

[waxwingslain]

Yes. The point is that, raising a complex number to the power of -1 does more than just flip the fraction, so your first equality isn't correct.

So it should be this?
z= ( -i [ 1 - i ] ) ^ -1
= ( 1 - i ) ^ -1
= i - 1

How did you come up with that first line?

Point is, if you're going to take the reciprocal of a complex number

(a + bi)/(c + di)

Then you can't just invert the fraction, like you can for real numbers. You use the formula

[(ac + bd) + (-ad + bc)i] / [c^2 + d^2]

which gives you (i - 1) / 2. Derivation is at http://www.clarku.edu/~djoyce/complex/div.html if you want to see it.

 

[lostclause]

Yes. The point is that, raising a complex number to the power of -1 does more than just flip the fraction, so your first equality isn't correct.

So it should be this?
z= ( -i [ 1 - i ] ) ^ -1
= ( 1 - i ) ^ -1
= i - 1

How did you come up with that first line?

Point is, if you're going to take the reciprocal of a complex number

(a + bi)/(c + di)

Then you can't just invert the fraction, like you can for real numbers. You use the formula

[(ac + bd) + (-ad + bc)i] / [c^2 + d^2]

which gives you (i - 1) / 2, the answer in your book. Derivation is at http://www.clarku.edu/~djoyce/complex/div.html if you want to see it.

Changed it into ( [ 1 / i ] x [ 1 - i ] ) ^ -1 instead of the original ( [ 1 - i ] / i ) ^ -1
since you said 1 / i = -i I just changed that but it gave me the same thing.
Anyway, I can see where I went wrong with this one, thanks for your help.