MrDudeMan
Lifer
- Jan 15, 2001
- 15,069
- 92
- 91
Originally posted by: TuxDave
Originally posted by: stan394
so what's the answer?
Originally posted by: TuxDave
Originally posted by: stan394
so what's the answer?
Originally posted by: coomar
never fails to amuse
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.
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
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.
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.
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.
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
Originally posted by: coomar
never fails to amuse
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
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
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.
Originally posted by: stan394
I think we got spidered...
I asked a Math PhD and he said there is no solution either.
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?
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?
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?