Complex sign. I mean sine!

Algebra Level 4

$\large \frac{\pi}{A} - i \ln {\left(B+\sqrt{C}\right)}$

There is no real value for $\sin^{-1}(2)$ but it has a complex one. And it is of the form as described above where, $A$ , $B$ and $C$ are positive integers. Evaluate $A+B+C$ .

The answer is 7.

2 solutions

Kazem Sepehrinia
Aug 16, 2015

Let $x=\sin^{-1} 2$ , so $\sin x=2$ or $\frac{e^{ix}-e^{-ix}}{2i}=2$ Let $y=e^{ix}$ and get a quadratic $y-\frac{1}{y}=4i \\ y^2-4iy-1=0 \\ y=i(2\pm\sqrt{3})$ Now for evaluating $x$ we have $e^{ix}=i(2\pm\sqrt{3}) \\ ix= \ln(i(2\pm\sqrt{3})) \\ x=-i \ln(i(2\pm\sqrt{3}))$ Positive branch must be chosen, thus $\begin{array}{c}\\ x&=-i \ln i -i \ln(2+\sqrt{3}) \\ &=-i \frac{i\pi}{2}-i \ln(2+\sqrt{3}) \\ &=\frac{\pi}{2}-i \ln(2+\sqrt{3}) \end{array}$

What critteria did you used for choose the positive branch? Thanks.

Daniel Rabelo - 5 years, 9 months ago

Thanks for your solution

Chew-Seong Cheong - 5 years, 10 months ago

No problem sir!

Kazem Sepehrinia - 5 years, 10 months ago

An other solution:

We put the point M(a,b) such as ; sin(-1).(2) = a + ib

so sin(a+ib) = 2

so sin(a)cos(ib)+sin(ib)cos(a) = 2

.And we know that: sinh(b)=i.sin(ib) and cosh(b)=cos(ib) (by definition)

so sin(a)cosh(b) - i.sinh(b)cos(a) = 2

so the imaginary part equal to zero, This means that either sinh(b) or cos(a) must be zero so: sinh(b) = 0 or cos(a) = 0

but if sinh(b) were zero, that would mean b=0, which would mean x was real. Since we know that is not the case so: cos(a)=0. Which means that a = pi/2

We go back to the real part : sin(a)cosh(b) = 2

And sin(a) = 1 cause a = pi/2

so cosh(b) = 2

And we know that : cosh(b)= ( exp(b) + exp(-b) ) / 2. (by definition)

so after solving this equation ; ( exp(b) + exp(-b) ) / 2 = 2

we get ; exp(b) = 2 + square root (3) so b = ln( 2 + square root (3) )

so a = pi/2 and b = ln( 2 + square root (3) )

then we get sin(-1).(2) = pi/2 + ln( 2 + square root (3) )

so a+b+c=7!

Abdelghani ßoris - 5 years, 10 months ago

My question, why $\\ \sin{x}= \frac{e^{ix}-e^{-ix}}{2i} \\$ and why $\\ \ln{i}= \frac{i \pi}{2} \\$ thanks

Ben Habeahan - 5 years, 9 months ago

For the sin it's known as Euler trigonometric formula, and for the second we know that i= exp( i*(pi/2) ) so ln(i) = ln( exp( i*(pi/2) ) ) so ln(i) = i*(pi/2)

Abdelghani ßoris - 5 years, 9 months ago

Thanks for your answers.

Ben Habeahan - 5 years, 9 months ago

Consider a case $arcsin2 =$ $\pi\over2$ $- i\ln{2}$ where $i = \sqrt -1$ now $2= \sin(\pi/2 -i\ln2$ so $\sin(\pi/2 -i\ln2$ is equivalent to imaginary part of $e^{i\pi/2 + \ln2}$ which is $2e^{i\pi/2}$ which gives us imaginary part $=2$ but this is no where close to your solution. If i am going wrong anywhere please tell.

neelesh vij - 5 years, 6 months ago

$\arcsin 2 = \dfrac{\pi}{2} - i \ln(2+\color{#D61F06}{\sqrt{3}})$

Chew-Seong Cheong - 5 years, 6 months ago

Lu Chee Ket
Oct 15, 2015

I think Pi/2 + j Ln (2 + Sqrt (3)) or Pi/2 - j Ln (2 - Sqrt (3)) is for Acosh (2) = 1.316957896924816708625046347308 of positive value only despite graph for forward function, for a one to one mapping to inverse function likes Asin (1) or Asin (2). Just like x = Sqrt (2) for x^2 = 2 but Sqrt (2) <> - Sqrt (2).

We can't make Sqrt (C) = - Sqrt (3) strictly. Therefore, the question should avoid mapping of Acosh (2) to -1.316957896924816708625046347308 or Asin (2) to Pi/2 - j Ln (2 + Sqrt (3)).

Sin p Cosh q + j Cos p Sinh q = 2 + j 0

q <> 0 therefore Cos p = 0 => p = Pi/ 2;

Sin Pi/ 2 Cosh q = 2 => q = Acosh (2) = Ln (2 + Sqrt (3)) or - Ln (2 - Sqrt (3)) only! {Your calculator shall give you only 1.316957896924816708625046347308 because this is a 1 to 1 mapping.

But we are talking about $\arcsin(2) = \dfrac{\pi}{2} \color{#D61F06}{-} i \ln (2+\sqrt{3})$ and not $\dfrac{\pi}{2} \color{#D61F06}{+} i \ln (2+\sqrt{3})$

You can add "\ (" before and "\ )" after (without the space) of your formulas to appear nicely in LaTex here. You can see what we have keyed in in LaTex by place your mouse cursor over the formulas.

Chew-Seong Cheong - 5 years, 8 months ago

Complex sign. I mean sine!

The answer is 7.

2 solutions

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