The most interesting math problem I've seen in a long time

[MrDudeMan]

Originally posted by: TuxDave

Originally posted by: stan394
so what's the answer?

 

[kitkat22]

I was curious and decided to plug this into the TI-83 equation solver and it came up with -9.932832623312e98. I don't believe this actually works, but according to TI...

 

[yhelothar]

Originally posted by: coomar
never fails to amuse

I examined that for a few minutes when I first saw that but figured it out.
If you look at the coordinates, the slope the second triangle looks slightly different from the top triangle - thus making up for the the lost space in the missing square.

 

[JohnCU]

I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

 

[kitkat22]

I figured it out. Apparentely -9.932832623312e98 is the number where the calculator does the rounding for you. It's like 3/3.3333333333333 in reality the number is .899999999999, but the calc answers .9.

 

[Soccer55]

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

Is ArcTan defined on the complex numbers? More specifically, is ArcTan(j) not undefined?

-Tom

 

[JohnCU]

Originally posted by: Soccer55

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

Is ArcTan defined on the complex numbers? More specifically, is ArcTan(j) not undefined?

-Tom

no, i do not think so. i'm just grasping at any possible solution right now.

 

[Soccer55]

Originally posted by: JohnCU

Originally posted by: Soccer55

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

Is ArcTan defined on the complex numbers? More specifically, is ArcTan(j) not undefined?

-Tom

no, i do not think so. i'm just grasping at any possible solution right now.

It looks good to me except for the ArcTan thing. I'm inclined to think that this isn't possible.....unless we're missing some information or something like that.

-Tom

 

[Yomicron]

The answer is pretty obvious:

z = 0 - 1

e^z = e^0 - 1 = 0

 

[MrDudeMan]

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

lol nice. im not sure if that is acceptable in complex analysis but it looks good anyway.

ANSWER PLEASE YOU FLAMER!

i asked all of my professors today and they all agreed it isnt possible. several of them tried to work something up but they all came up with no solution, which is exactly what i think the answer is - no solution.

if there is a solution, then it is some sort of trick or BS...just incase it is real math, however, please post it and prove me and several professors wrong.

 

[Soccer55]

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

After looking over this again, shouldn't that be Tan(y) = -1/j since 1/(-j) = siny/cosy? Which would then make your answer y = ArcTan(-1/j)?

-Tom

 

[JohnCU]

Originally posted by: Soccer55

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

After looking over this again, shouldn't that be Tan(y) = -1/j since 1/(-j) = siny/cosy? Which would then make your answer y = ArcTan(-1/j)?

-Tom

1/(-j) is the same thing as j

 

[TheoPetro]

Originally posted by: coomar
never fails to amuse

lol you SOB haha. i spent 5 min staring at that computing the areas over and over one was 32.5 the other was 32. then i realized the overall shape WASNT A TRIANGLE.....that was very deceitful, but amusing once you figure it out. (.375 and .4 for slopes are too dang similar)

 

[chuckywang]

Originally posted by: JohnCU

Originally posted by: Soccer55

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

After looking over this again, shouldn't that be Tan(y) = -1/j since 1/(-j) = siny/cosy? Which would then make your answer y = ArcTan(-1/j)?

-Tom

1/(-j) is the same thing as j

I can tell you're an engineer for using j. LOL.

 

[Soccer55]

Originally posted by: JohnCU

Originally posted by: Soccer55

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

After looking over this again, shouldn't that be Tan(y) = -1/j since 1/(-j) = siny/cosy? Which would then make your answer y = ArcTan(-1/j)?

-Tom

1/(-j) is the same thing as j

heh.....oops.....shoulda known that

-Tom

 

[MrDudeMan]

Originally posted by: Yomicron
The answer is pretty obvious:

z = 0 - 1

e^z = e^0 - 1 = 0

Wrong


somehow you made z equal to 0 - 1 and got the -1 outside of the power of e. you might need to rethink your math.

 

[MrDudeMan]

Originally posted by: chuckywang

Originally posted by: JohnCU

Originally posted by: Soccer55

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

After looking over this again, shouldn't that be Tan(y) = -1/j since 1/(-j) = siny/cosy? Which would then make your answer y = ArcTan(-1/j)?

-Tom

1/(-j) is the same thing as j

I can tell you're an engineer for using j. LOL.

i like using j instead of i regardless of engineering

 

[stan394]

I think we got spidered...

I asked a Math PhD and he said there is no solution either.

 

[stan394]

Originally posted by: stan394
I think we got spidered...

I asked a Math PhD and he said there is no solution either.

this is what my friend said

"could the answer be some kind of joke?

in mathematics, the exponential function can be generalized in many
ways, once you take its defintion to be "e^z = 1+ sum of z^n/(n!) over
all n>=1". for example, z can be a complex number, a square matrix, or
anything one can take powers of (so that the above infinite sum makes
sense). these generalizations are actually useful, e.g. exponential of
a square matrix helps solving systems of differential equations. but
i've never seen a situation in which one has e^z=0. one reason is
this. as long as the things (complex numbers, square matrices) you are
considering have addition and multiplication satisfying some usual
properties (like a(b+c)=ab+ac), one has e^z x e^(-z) = 1. so if e^z=0
is true (and is not a joke), z must be something with an unusual
multiplication rule, but it's not impossible."

 

[DaShen]

Originally posted by: jonessoda
ln 0. Easy.

funny.

 

[TheRyuu]

So, would this be true? :

e^z=0
zlog(e)=log0
z=log(0/e)???
z=log(0)-log(e)???

right? Or totally wrong?

 

[Soccer55]

Originally posted by: wizboy11
So, would this be true? :

e^z=0
zlog(e)=log0
z=log(0/e)???
z=log(0)-log(e)???

right? Or totally wrong?

log(0)/log(e) != log(0/e)

-Tom

 

[stan394]

Originally posted by: wizboy11
So, would this be true? :

e^z=0
zlog(e)=log0
z=log(0/e)???
z=log(0)-log(e)???

right? Or totally wrong?

log(0) is undefined, right?

 

[Soccer55]

Originally posted by: stan394

Originally posted by: wizboy11
So, would this be true? :

e^z=0
zlog(e)=log0
z=log(0/e)???
z=log(0)-log(e)???

right? Or totally wrong?

log(0) is undefined, right?

You speak the truth.

-Tom

 

[Jack Ryan]

http://www.csh.rit.edu/~pat/math/quickies/epii/

http://mathforum.org/library/drmath/view/51921.html