Complex Number Help!

[Damo17]

One more for the night that I gave up on yesterday.

If , determine the values of and such that .

[Mao]

Therefore,
as required.

on a side note, it is true that , not only limited to even integers, since the definition of the imaginary number is

that was enwiabe's reasoning. I personally don't agree with it that much, as when b or c is a fraction is generally valid only if a is positive.

it is analogous to this: http://en.wikipedia.org/wiki/False_proof#Version_3

i would probably have done this:



but for Damo's sake, it was logical enough

[Mao]

One more for the night that I gave up on yesterday.

If , determine the values of x and y such that .

equating real and imaginary:

[Collin Li]

Equating real and imaginary parts:

--{1}

--{2}

From {2}:

Substituting into {1}:



Using the quadratic formula with respect to :





(rejecting the negative solution -- since and are real numbers)





or

[Collin Li]

When using the polar form method to obtain roots is ugly and cumbersome, sometimes cartesian expansion will do the job more efficiently (especially if its just square roots).

You can do it with higher powers to obtain some fairly cool trigonometric equalities too.

[Damo17]

Thanks Mao and Coblin. What great, helpful and kind people we have on here. Thanks alot!

When using the polar form method to obtain roots is ugly and cumbersome, sometimes cartesian expansion will do the job more efficiently (especially if its just square roots).

You can do it with higher powers to obtain some fairly cool trigonometric equalities too.

I've just finished Conjugates and division of complex numbers.
I'm going to learn Polar form of Complex Numbers tomorrow and the basic operations of them.

[Ahmad]

Here is another way to do it,

If I can find the complex form of point E, I can readjust the length of this complex number so that it's and I'll be done.

AB was extended so that angle ABE = ADC. Simple arithmetic shows that . Therefore is isosceles therefore, .

Therefore, is similar to , this means,

, and since we have

Therefore adjusting to make it have length gives , and we mustn't forget the other negative solution, therefore .

[Umesh]

wow

[bigtick]

on a side note, it is true that , not only limited to even integers, since the definition of the imaginary number is

It is true only for even natural numbers.

[Mao]

did you try using any non-even decimal numbers?

n=0.24692 works
n=-19.6 also works...
works also...

[bigtick]

I was referring to the original question. What I meant to say was, it is not true for odd natural numbers. You said it is not only limited to even integers. You have to think complex consistently. I think you have missed the essence of the question.

[Mao]

I was referring to the original question. What I meant to say was, it is not true for odd natural numbers. You said it is not only limited to even integers. You have to think complex consistently. I think you have missed the essence of the question.

please explain how it is not true for odd natural numbers.

[bigtick]

There is another value for (-1)^(1/2) besides i. Think complex.

[Mao]

the notation of is equivalent to , which is the principle square root.

if the question did not mean the principle square root, it would have been presented as

[Collin Li]

Problems arise for the odd powers when taking