# Complex numbers, inverse trig and hyperbolic

• Liquidxlax
In summary, the student is trying to solve for z in a problem, but is getting confused because the professor is using complex notation. They eventually figure out what to do and get the same result as what was shown in a previous post.
Liquidxlax

## Homework Statement

edit* It says Verify the formulas in problems

arcsin(z) = -iln(iz ±sqrt(1-z^2))

arccos(z) = iln(z ±sqrt(1-z^2))tanh-1z = (1/2)ln((1+z)/(1-z))

## The Attempt at a Solution

yeah, my prof just threw it at us, all i have is nothing... absolutely nothing. I don't know why he does this to us.

The best i can think of isz= sin(-iln(iz ±sqrt(1-z^2)))

z = (eln(iz ±sqrt(1-z^2)) + e-ln(iz ±sqrt(1-z^2)))/2i((iz ±sqrt(1-z^2)) + (iz ±sqrt(1-z^2)-1)/2i

but i doubt that is right

You have really got to get squared with the complex logarithm and stop using the ln symbol to denote it. Use ln only when referring to the real log function. Use log to refer to the complex log function.

So we know:

$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=w$$

Now, how do you solve for z in that? I'll show you some of the steps. You try to fill in the blanks if you want:

$$2iw=e^{iz}-\frac{1}{e^{iz}}$$

$$2iwe^{iz}=e^{2iz}-1$$

$$e^{iz}=\frac{2iw+\sqrt{-4w^2+4}}{2}$$

$$iz=\log(iw+\sqrt{1-w^2})$$

Go through that one, then try arccos(z) the same way.

Keep in mind that's a complex log function: $\log(z)=\ln|z|+i(\theta+2n\pi)$ and note I do not use $\pm$ for the root. In Complex Analysis, it's implied to have two values. So the arcsin function is infinitely-valued and for each one of those n-values of the log function, the square root has two different values. So if I asked what is the arcsin(z) when n=10,11, there would be four values of the arcsin function for those two values of n.

Hi Liquidxlax!

(have a square-root: √ and a theta: θ and try using the X2 icon just above the Reply box )
Liquidxlax said:
Verify the formulas in problems

arcsin(z) = -iln(iz ±sqrt(1-z^2))

eugh! … let's keep this simple

whenever I see √(1 - z2), I immediately want to do a trig subsitution.

So forget about the LHS, and just put z = sinθ in the RHS …

what do you get?

tiny-tim said:
Hi Liquidxlax!

(have a square-root: √ and a theta: θ and try using the X2 icon just above the Reply box )eugh! … let's keep this simple

whenever I see √(1 - z2), I immediately want to do a trig subsitution.

So forget about the LHS, and just put z = sinθ in the RHS …

what do you get?

i don't think that will help, but i did what jack said, yet i did not get the right answerfor arcsin

(iz +- sqrt(1-z^2))^2 = Log(iz +- sqrt(1-z^2)) somehow

Liquidxlax said:
i don't think that will help, but i did what jack said, yet i did not get the right answerfor arcsin

(iz +- sqrt(1-z^2))^2 = Log(iz +- sqrt(1-z^2)) somehow

$$w=\frac{e^{iz}-e^{-iz}}{2i}$$

and solving for z like I showed above, do you not get the same expressions I obtain in the above post?

Last edited:
jackmell said:

$$w=\frac{e^{iz}-e^{-iz}}{2i}$$

and solving for z like I showed above, do you not get the same expressions I obtain in the above post?

nope... but isn't it arcsinz = w, or should i ignore the rhs and just say

arcsinz = w => z=sinw => z = (e^iw - e^-iw)/2i. I mean they're very very stict with notation and variables i can't be changing them randomly even if i do say let z = w or something, they still whine and complain.

being my typical in a rush self, i slowed down and did what you said Jackmell, and it worked just fine

## 1. What are complex numbers?

Complex numbers are numbers that are composed of both a real part and an imaginary part. They are typically written in the form a + bi, where a is the real part and bi is the imaginary part. The imaginary part is a multiple of the imaginary unit i, which is defined as the square root of -1.

## 2. How are complex numbers used in science?

Complex numbers are used in various fields of science, such as physics, engineering, and mathematics. They are often used to represent physical quantities that have both magnitude and direction, such as electric fields and alternating currents. They are also used in signal processing and control systems.

## 3. What is the inverse trigonometric function?

The inverse trigonometric function, also known as the arc trigonometric function, is the inverse of a trigonometric function such as sine, cosine, or tangent. It takes a ratio of two sides of a right triangle as input and returns the angle that would produce that ratio. It is denoted by the prefix "arc" or by using the notation sin-1 or arcsin.

## 4. What is hyperbolic function?

Hyperbolic functions are a set of mathematical functions that are analogous to trigonometric functions. They are defined in terms of the hyperbolic sine, cosine, tangent, and their inverses. They are commonly used in fields such as physics, engineering, and economics to model various phenomena, such as growth and decay.

## 5. How are inverse trigonometric and hyperbolic functions related?

Inverse trigonometric and hyperbolic functions are related in that they both involve finding the angle or input value that would produce a given output. The main difference is that inverse trigonometric functions are used to solve for angles in a right triangle, while hyperbolic functions are used to solve for input values in a hyperbolic function. Additionally, they both have similar properties and identities, such as the inverse trigonometric identity sin(arcsin x) = x and the hyperbolic identity sinh(arcsinh x) = x.

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