determine arcsinh(2+i) in form of a+bi
thanks
I'm sure you know that
arcsinh(z) = log(z+√(1+z^2))
so,
arcsinh(2+i) = log(2+i + √(4+2i))
= log(4.058 + 1.486i)
Now log(z) = log(x+yi)
= log(√(x^2+y^2) + i arctan(y/x)
So, plug in your z value above and crank it out.
google arcsinh to get the definition I gave
google complex logarithm to find how to evaluate it.
Surely you have enough keywords to do some research yourself. I didn't know that stuff either, but I know how to search for it.
To determine the inverse hyperbolic sine (arcsinh) of a complex number, we can follow these steps:
Step 1: Let's assume the given complex number is z = 2 + i.
Step 2: Express z in terms of the exponential form, z = re^(iθ), where r is the modulus (distance from the origin) and θ is the argument (angle with the positive real axis).
To find r, we calculate the modulus of z:
r = |z| = sqrt(2^2 + 1^2) = sqrt(5) = √5
To find θ, we calculate the argument of z:
θ = arg(z) = arctan(1/2) = 0.4636 radians
Step 3: Now, we can express z in exponential form:
z = √5 * e^(i * 0.4636)
Step 4: To find arcsinh(z), use the following formula:
arcsinh(z) = ln(z + sqrt(z^2 + 1))
Substituting the value of z we calculated in step 3:
arcsinh(2 + i) = ln((√5 * e^(i * 0.4636)) + sqrt((√5 * e^(i * 0.4636))^2 + 1))
This gives us the result in the form of natural logarithm. If you want to express it in a+bi form, you can convert it by using the complex logarithm properties.