What Are the Roots of sinh(z) = 1/2?

[whizzkid]

Homework Statement

Find all the roots of sin h(z) = 1/2

2. The attempt at a solution

sin h(z) = [1/2](e^z - e^-z) = 1/2
=> e^z -e^-z = 1
=> e^2z - e^z - 1 = 0 {multiplied e^z bothsides}
this is a quadratic equation in e^z using quadratic formula,

e^z = [1+- sqrt(5)]/2

taking 'ln' on bothsides

z = ln[{1+- sqrt(5)}/2]

And after this step I am confused, what to do next and how to find all the roots?

can someone help me out in this?

 

[dextercioby]

Just remember that

$$e^{2k\pi i} =1 \ , \forall k\in\mathbb{Z}$$

Daniel.

 

[whizzkid]

i can think of
z = ln(1/2 +-(sqrt(5)/2)
z = ln(1.61) and z = ln(-0.61)

and ln(-ive) is not defined.

:(

 

[dextercioby]

No way is it true that ln(a+b)=ln a +ln b.

Daniel.

 

[whizzkid]

yupp u r right!

 

[cristo]

Nowhere on that page does it state that ln(a+b)=lna+lnb.

The correct result, which is stated, is ln(ab)=lna+lnb

 

[whizzkid]

omg... cristo... I got ur point.I m making a blunder actually ...ok
i come up wid sumthing new and edited my aforementioned post!

 

[whizzkid]

z = ln(1/2 +-(sqrt(5)/2)

z = ln(1+sqrt(5)/2) and z = ln(1-sqrt(5)/2)
z = ln(3.24)-ln(2)+2k*pi*i and z = ln(1.24)-ln(2)+2k*pi*i

0.48+2k*pi*i and -0.48+2k*pi*i

we can say +-1/2 + 2k*pi*i

is this solution correct?

 

[dextercioby]

You can't approximate to 1/2. It's actually better if you leave it in the original form, the one still containing natural logs.

Daniel.

 

[HallsofIvy]

z = ln(1/2 +-(sqrt(5)/2)

z = ln(1+sqrt(5)/2) and z = ln(1-sqrt(5)/2)
z = ln(3.24)-ln(2)+2k*pi*i and z = ln(1.24)-ln(2)+2k*pi*i

0.48+2k*pi*i and -0.48+2k*pi*i

we can say +-1/2 + 2k*pi*i

is this solution correct?

0.48 is not 1/2! Don't approximate.

By the way, it is sinh(x) not sin h(x). I first thought you meant sine of some function h(x)!