Help with complex numbers(locus) and hyperbolic functions

[CharlesL]

Homework Statement

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.

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 )

Homework Statement

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.

The Attempt at a Solution

I first subsituted [ e^x - e^-x / 2 ] into all the sinh available in the equation with their specific value of x.

9[ e^4x - e^-4x / 2 ] - 82[ e^3x - e^-3x / 2 ] + 9[e^2x - e^-2x/ 2 ] = 0

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

9e^4x - 9e^-4x- 82e^3x + 82e^-3x + 9e^2x - 9e^-2x = 0

Then I tried to separate e^4x to e⁴ . e^x and regroup the ones with e^x and e^-x

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

Then I multiplied both sides with e^x

[e^x]² = 1

e^x= 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

 

[Office_Shredder]

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

 

[CharlesL]

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 I am still lost. Maybe completing the squares wasnt the right step?

 

[xepma]

I'm confused with your notation...

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...

 

[CharlesL]

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.

 

[xepma]

Since when is the square root of 1/2 not a number? :)

 

[CharlesL]

Oh.. Hahahha! Alright, I guess I am done with Question 1 with the help from you and Office_Shredder. Thank you both again.

Would love some help with Question 2 though. :)

 

[nrqed]

Oh.. Hahahha! Alright, I guess I am 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 $$e^4 e^x$$ !

 

[CharlesL]

Thanks nrqed! Was looking for someone to ask whether if I could do that. Since I can't 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?

 

[Office_Shredder]

If you write $$e^{4x} = (e^{x})^4$$ etc., then you can make the substitution y=e^x 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.

 

[CharlesL]

Thank you Office_Shredder for your kind help. I tried to subsitute y=e^x and got the quartic equation which unfortunately I am 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 :(

 

[Office_Shredder]

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

 

[CharlesL]

Thank you office_shredder for your kind reply again. So from y⁶ = 1. Which also means y =1?

Therefore from y = e^x

1 = e^x
ln 1 = x ln e
x = 0?

 

[Office_Shredder]

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

 

[CharlesL]

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. :)

 

[HallsofIvy]

Homework Statement

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.

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 )

Homework Statement

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.

The Attempt at a Solution

I first subsituted [ e^x - e^-x / 2 ] into all the sinh available in the equation with their specific value of x.

9[ e^4x - e^-4x / 2 ] - 82[ e^3x - e^-3x / 2 ] + 9[e^2x - e^-2x/ 2 ] = 0

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

9e^4x - 9e^-4x- 82e^3x + 82e^-3x + 9e^2x - 9e^-2x = 0

Then I tried to separate e^4x to e⁴ . e^x and regroup the ones with e^x and e^-x

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

Then I multiplied both sides with e^x

[e^x]² = 1

e^x= 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