## Help with complex numbers(locus) and hyperbolic functions

1. The problem statement, all variables and given/known data
Question(1) : Find the Cartesian equation of Re[ z - i / z + 1 ] = 0. If the locus is a circle, give its radius and the coordinates of its center.

3. The attempt at a solution
Workings : So I attempted to solve the problem and my workings are as below

... Since Re = Real part,

Let z = x + iy

Re[ x + iy - i / x + iy + 1 ] = 0

x/x+1 = 0

x = 0

Right here I am assuming that the locus is at all the points of the line x=0.

NEXT, to obtain the radius and the coordinates of the center of the locus(circle),

[modulus] z - i/ z + 1 [modulus] = 0
[modulus] z - i [modulus] = 0
[modulus] x + iy - i [modulus]= 0
square root[ (x - 0)2 + (y - 1)2 ] = 0
(x - 0)2 + (y - 1)2 = 0

Therefore, the radius of the circle is 0 and the coordinates of the center is ( 0, 1 )

1. The problem statement, all variables and given/known data
Question (2) : Obtain all the real solutions of the following equation: 9 sinh 4x - 82 sinh 3x + 9 sinh 2x = 0 . Show all your derivations.

3. The attempt at a solution
I first subsituted [ ex - e-x / 2 ] into all the sinh available in the equation with their specific value of x.

9[ e4x - e-4x / 2 ] - 82[ e3x - e-3x / 2 ] + 9[e2x - e-2x/ 2 ] = 0

I multiply the whole equation by 2 and decided to multiply the integer outside of each boxes,

9e4x - 9e-4x- 82e3x + 82e-3x + 9e2x - 9e-2x = 0

Then I tried to separate e4x to e4 . ex and regroup the ones with ex and e-x

[ 9e4- 82e3 + 9e2 ] ex = [ 9e4 - 82e3 + 9e2 ] e-x

Then I multiplied both sides with ex

[ex]2 = 1

ex= square root + of 1 (chosed only +ve value as the question mentioned about real solutions)

Then I applied ln

x ln e = ln square root of 1

x = 0

Any comments would be a great help and much appreciated. Thank you in advance and have a nice day.

Regards
Charles
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 Blog Entries: 1 Recognitions: Homework Help For the first one: Re(a/b) =/= Re(a)/Re(b) in general e.g. (1+i)/(2+i) = (1+i)(2-i)/5 = (3+i)/5 which has real part 3/5, not 1/2 Furthermore, a line is never a circle
 Thanks for the heads up Office_Shredder Well then I think I should then make my first step by multiplying a conjugate of the denominator (z-i)/(z+1) = 0 [( x2 + x + y2 - y ) + i ( y-x-1)] / x2 + 2x + 1 + y2 = 0 Then i tried completing the squares (with the Real part = 0) to get the equation for circle (x-a)2 + (y-b)2 = r2 I end up with [x2 + x + (-1/2)2 ] - (-1/2)2 + [y2 - y + (1/2)2] - (1/2)2 = 0 [x -(-)(-1/2)]2 + [y-1/2]2 = (-1/2)2 + (1/2)2 I end up with my radius being 0 and coordinates as (1/2,1/2). There is no circles with a 0 radius so Im still lost. Maybe completing the squares wasnt the right step?

## Help with complex numbers(locus) and hyperbolic functions

Do you mean: Re[ (z-i)/(z+1)] or Re[z - (i/z) + 1] or maybe even something else...

I'm guessing you mean the first one. In that case, you're on the right track. You made a small error in finishing the square of the x-dependent part (a minus sign). Furthermore, the radius of your circle is not zero. The squared numbers on the right hand side do not add up to 0...
 xepma I apologise for the confusion that I have caused. Yes I meant (z-i)/(z+1) = 0. After you have pointed out my error, this is what I've got and please let me know if it is the correct way of doing it. [x - (-1/2)]2 + [y-1/2]2 = (-1/2)2 + (1/2)2 As for the radius, if I were to sum up the right hand side, the result would be (1/4) + (1/4) = 1/2 From the general equation of the circle (x-a)2 + (y-b)2 = r2 the value of r should be squared. By summing up the right hand side after squaring each fraction, it would not produce a squared value. Please correct me if I am wrong. Thank you for your help.
 Since when is the square root of 1/2 not a number? :)
 Oh.. Hahahha! Alright, I guess Im done with Question 1 with the help from you and Office_Shredder. Thank you both again. Would love some help with Question 2 though. :)

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 Quote by CharlesL Oh.. Hahahha! Alright, I guess Im done with Question 1 with the help from you and Office_Shredder. Thank you both again. Would love some help with Question 2 though. :)
Watch out!
$$e^{4x}$$ is not equal to $$e^4 e^x$$ !
 Thanks nrqed! Was looking for someone to ask whether if I could do that. Since I cant separate the 4 and the x, what do you reckon I should do for my next step? Or maybe even my first step was wrong?
 Blog Entries: 1 Recognitions: Homework Help If you write $$e^{4x} = (e^{x})^4$$ etc., then you can make the substitution y=ex to get a quartic equation in y. You should be able to reduce that to something you can solve for y for, and then get x from y.
 Thank you Office_Shredder for your kind help. I tried to subsitute y=ex and got the quartic equation which unfortunately Im not very familiar with. 9y4 - 9y-4 - 82y3 + 82y-3 + 9y2 - 9y-2 = 0 I didnt know what to do with the negative powers so I mulitplied the equation with y4 to eliminate the negatives 9y8 - 82y7 + 9y6 - 9y2 + 82y - 9 = 0 which got me lost :(
 Blog Entries: 1 Recognitions: Homework Help Look at how you can pair stuff up: 9y8 - 9y2 82y - 82y7 9y6-9 So if y6 = 1, then you get that each of those terms is zero, and hence you've found six (complex) solutions to that polynomial
 Thank you office_shredder for your kind reply again. So from y6 = 1. Which also means y =1? Therefore from y = ex 1 = ex ln 1 = x ln e x = 0?
 Blog Entries: 1 Recognitions: Homework Help No. Any polynomial of degree n has exactly n complex roots. So there are actually five complex roots of y6=1. Of course, these won't be relevant since you're only searching for real solutions to the original equation, but it guides us to realizing that there also are 8 roots of the degree eight polynomial. We find the other two by factoring: We now know y6-1 divides 9y8 - 82y'7 + 9y6 - 9y2 + 82y - 9 so we factor to get: 9*(y6-1)*(y2 - 9y + 1) = 0 So you should be able to find the other two roots from here
 Yes I got the other two roots from your guide office_shredder. Really appreciate all the help from you guys. This is a really educative and helpful forum. Thank you so much guys. :)

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 Quote by CharlesL 1. The problem statement, all variables and given/known data Question(1) : Find the Cartesian equation of Re[ z - i / z + 1 ] = 0. If the locus is a circle, give its radius and the coordinates of its center. 3. The attempt at a solution Workings : So I attempted to solve the problem and my workings are as below ... Since Re = Real part, Let z = x + iy Re[ x + iy - i / x + iy + 1 ] = 0 x/x+1 = 0
How did you get this? If z= x+ iy then
$$Re\frac{z-i}{z+ 1}= Re\frac{x+ i(y-1)}{(x+1)+ iy}$$
multiplying numerator and denominator by (x+1)- iy, we get
$$\frac{x^2+ x- (y^2+y)}{(x+1)^2+ y^2}+ i (.... )$$

 x = 0 Right here I am assuming that the locus is at all the points of the line x=0. NEXT, to obtain the radius and the coordinates of the center of the locus(circle),
You just said that you were assuming it was a line. How did it become a circle in the next sentence?

 [modulus] z - i/ z + 1 [modulus] = 0 [modulus] z - i [modulus] = 0 [modulus] x + iy - i [modulus]= 0 square root[ (x - 0)2 + (y - 1)2 ] = 0 (x - 0)2 + (y - 1)2 = 0 Therefore, the radius of the circle is 0 and the coordinates of the center is ( 0, 1 ) 1. The problem statement, all variables and given/known data Question (2) : Obtain all the real solutions of the following equation: 9 sinh 4x - 82 sinh 3x + 9 sinh 2x = 0 . Show all your derivations. 3. The attempt at a solution I first subsituted [ ex - e-x / 2 ] into all the sinh available in the equation with their specific value of x. 9[ e4x - e-4x / 2 ] - 82[ e3x - e-3x / 2 ] + 9[e2x - e-2x/ 2 ] = 0 I multiply the whole equation by 2 and decided to multiply the integer outside of each boxes, 9e4x - 9e-4x- 82e3x + 82e-3x + 9e2x - 9e-2x = 0 Then I tried to separate e4x to e4 . ex and regroup the ones with ex and e-x [ 9e4- 82e3 + 9e2 ] ex = [ 9e4 - 82e3 + 9e2 ] e-x Then I multiplied both sides with ex [ex]2 = 1 ex= square root + of 1 (chosed only +ve value as the question mentioned about real solutions) Then I applied ln x ln e = ln square root of 1 x = 0 Any comments would be a great help and much appreciated. Thank you in advance and have a nice day. Regards Charles