Asked by hang
determine arcsinh(2+i) in form of a+bi
thanks
thanks
Answers
Answered by
Steve
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.
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.
Answered by
Steve
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.
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.
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