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How to calculate the inverse hyperbolic functions of a complex number?

To compute the inverse hyperbolic functions of a complex number, we can apply specific formulas designed for these functions. The definitions of the inverse hyperbolic functions are as follows:

arcsinh(z)=ln(z+z2+1)\text{arcsinh}(z) = \ln\left(z + \sqrt{z^2 + 1}\right) arccosh(z)=ln(z+z21)\text{arccosh}(z) = \ln\left(z + \sqrt{z^2 - 1}\right) arctanh(z)=12ln(1+z1z)\text{arctanh}(z) = \frac{1}{2} \ln\left(\frac{1 + z}{1 - z}\right)

To find the inverse hyperbolic functions of a complex number, first express the complex number in terms of its real and imaginary parts, specifically as z=x+iyz = x + iy. Then, substitute xx and yy into the appropriate formula for the desired inverse hyperbolic function.

Example Calculations

1. Finding arcsinh(2+3i)\text{arcsinh}(2 + 3i)

We can compute this as follows:

arcsinh(2+3i)=ln(2+3i+(2+3i)2+1)\text{arcsinh}(2 + 3i) = \ln\left(2 + 3i + \sqrt{(2 + 3i)^2 + 1}\right)

Calculating the square, we have:

(2+3i)2=4+12i9=5+12i(2 + 3i)^2 = 4 + 12i - 9 = -5 + 12i

Thus,

(2+3i)2+1=4+12i(2 + 3i)^2 + 1 = -4 + 12i

Now we compute the square root:

4+12i\sqrt{-4 + 12i}

Next, we substitute back into our equation:

arcsinh(2+3i)=ln(2+3i+4+12i)\text{arcsinh}(2 + 3i) = \ln\left(2 + 3i + \sqrt{-4 + 12i}\right)

This expression continues, but let’s highlight the main steps without diving into all intermediate calculations. The final expression will yield:

arcsinh(2+3i)=ln(2+3i+2i23i)\text{arcsinh}(2 + 3i) = \ln\left(2 + 3i + 2i \sqrt{2 - 3i}\right)

2. Finding arccosh(2+3i)\text{arccosh}(2 + 3i)

Similarly, we compute:

arccosh(2+3i)=ln(2+3i+(2+3i)21)\text{arccosh}(2 + 3i) = \ln\left(2 + 3i + \sqrt{(2 + 3i)^2 - 1}\right)

Calculating the square:

(2+3i)21=5+12i1=6+12i(2 + 3i)^2 - 1 = -5 + 12i - 1 = -6 + 12i

Now, we compute the square root:

6+12i\sqrt{-6 + 12i}

Substituting this back, we get:

arccosh(2+3i)=ln(2+3i+6+12i)\text{arccosh}(2 + 3i) = \ln\left(2 + 3i + \sqrt{-6 + 12i}\right)

This will also yield a complex logarithmic expression upon further simplification.

3. Finding arctanh(2+3i)\text{arctanh}(2 + 3i)

For this function, we start with:

arctanh(2+3i)=12ln(1+(2+3i)1(2+3i))\text{arctanh}(2 + 3i) = \frac{1}{2} \ln\left(\frac{1 + (2 + 3i)}{1 - (2 + 3i)}\right)

Simplifying the fractions gives:

=12ln(1+3i13i)= \frac{1}{2} \ln\left(\frac{-1 + 3i}{-1 - 3i}\right)

This can be further simplified:

=12ln(86i10)= \frac{1}{2} \ln\left(\frac{8 - 6i}{10}\right)

Finally, we can express this as:

arctanh(2+3i)=12ln(43i)\text{arctanh}(2 + 3i) = \frac{1}{2} \ln(4 - 3i)

In summary, calculating the inverse hyperbolic functions of complex numbers involves expressing the complex number in terms of its real and imaginary parts, followed by substituting these values into the appropriate formulas. Each calculation can lead to complex logarithmic forms that require careful manipulation and simplification.

Answered by: Dr. Lucas Harris
A-Level Maths Tutor
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