Messing With Mathematics

[Omkar™ 0]

I am currently studying in preliminary college, and we have “Complex Numbers” as a part of our syllabus. Its definition can be made as-

 

Z = a + i b (Where a,b are real numbers and i = √-1 )

 

Now, we have i = √-1 and hence i becomes an imaginary number.

 

But we have figured out a way to expand our imagination, and relate it to reality! Here goes, just some mathematical operations playing with the unity and,

 

i = √-1 (By definition)

 

Therefore, i^2 = -1

 

(i^2)^2 = (-1)^2 (Squaring both sides)

 

i^4 = 1 <--- How cool is that?!

 

Now reverting back,

 

i = 4√1 (4th root of 1)

 

And hence, i = 1

 

Now isn’t that interesting? Nothing done illegally, within the rules of mathematics and voila! We’ve torn apart the definition of Complex numbers! But there might me a hitch, something done wrong, can you spot it out? What is the catch in this?

 

It took a whole lot of time for me to type this thing in, but anything for a forum reply! But really, if that happened, I would be at par with!

 

P.S.:If you're wondering what that garbelled sign means, its the square root sign (√).

Edited September 5, 2006 by Omkar™ (see edit history)

[pyost 0]

While it all does seem properly done, I believe that the reason is quite simple.Let's do the same thing with another example.i = -2i^2 = 4i^4 = 16 /// squaring both sidesThat gives us this:i=4th sqrt(16), which equals 2. Or does it? These types of equasions [x=sqrt(a), a>=0] have two solutions. Those are [sqrt(a)] and [-sqrt(a)].That way you get two solution for i: 1 and -1 There's no catch, you were just playing with the wrong result

[Giniu 0]

Had this few years back on first hour of analitic functions... - remember that complex number are algebraically closed field and that means every equation of form x^n = y (in more general any a_nx^n + a_(n-1)x^(n-1 + ... + a_0 = 0 for a_n =/= 0) have always exactly n solutions... in that case (x^4 = 1) four solutions - so one of solution is 1... call it (1,0) and draw on plane where (x,y)=(1,0) usually is... now draw circle of size 1 in center of (0,0). And divide the circle into 4 parts... staring from (1,0)... and you get 4 numbers:(1,0) = 1, (0, 1) = i, (-1, 0) = -1, (0, -1) = -iand all those suits equation you gave... remember that multiplication in complex can be read as rotation and scaling of vectornow how to find all posible solutions to x^20 = 1? draw (1,0), circle, and divide it into 20 parts... you can explicite calculate them if you are able to convert number from standard mode into trygonometric mode: instead of z = a + ib you can writte z = |z|*(cos a + i sin a) and get same result - then you can easily multiplicate or divide a to get new numbers and turn mode back to other form if you want/can)and all this because complex are algebraically closed field :)edit--also remember - you can write that i^2 = -1 (this is how i is defined in most cases) but writting sqrt(i) isn't correct way for complex

Edited September 5, 2006 by Giniu (see edit history)

[mitchellmckain 0]

The polar representation of complex numbers is also very helpful in understanding thisFor z = x + iy you can calcultate r=sqrt(x^2+y^2) and t = tan^-1(y/x), just like when you are calculating the polar coordinates from the cartesian coordinates of a point. Then z = r e^(i t). Converting back to the cartesian form is easier using z = r cos t + i r sin t.Now when you look at the power functions you find something interesting.z^2 = r^2 e^(2 i t)In other words squaring a complex number squares the distance from the origin and doubles the angle.Likewisez^(1/2) = r^(1/2) e^( i t / 2)The square root gives square root the distance from the origin and half the angle.But now the problem is that the angles do not have a unique number but, for example, 0 degrees and 360 degrees are the same angle. It doesn't make much different when you are squaring because 720 degrees is the same as 360 degrees. But in the square root, half 360 degrees is 180 degrees which is not the same angle.Thus 1 = 1 e^(i 0) = 1 e^(i 360)and sqrt(1) = 1 e^(i 0) and 1 e^(i 180), that is both 1 and -1likewise since -1 = 1 e^(i 180) = 1 e^(i 540) from adding 360 to 180we get sqrt(-1) = 1 e^(i 90) and 1 e^(i 270) which is i and -iSo for a general root z^(1/n) of z = r e^(i t) we will get n answersr^(1/n) e^(i t/n), r^(1/n) e^(i (t+360)/n), r^(1/n) e^(i (t+720)/n),... r^(1/n) e^(i (t+(n-1) 360)/n).You can see that for z=1 this means r=1 and the for the root z^(1/n) you have n possibilities which are the n equally spaced points on the unit circle including 1 as someone mentioned already.But in fact you can see this is true of any complex number on the unit circle.take for example -1 for which r = 1 and t = 180 degreesThen the fourth root gives e^(i 45), e^(i 135), e^(i 225), e^(i 315)or in cartesian form (i+1)/sqrt(2), (i-1)/sqrt(2), (-i-1)/sqrt(2), (1-i)/sqrt(2)Checking the first root means squaring (i+1)/sqrt(2) twice(i+1)/sqrt(2) times (i+1)/sqrt(2) = (-1 + 2i + 1)/2 = iand clearly i squared is -1.