express ln(sinh^-1(1+j) in form of a+jb

Question

help plz no ideal
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Answer 1

recall that
sinh^-1(z) = ln(z+√(z^2+1))

so, find that, and then convert to polar form. Finally,

ln z = ln(r e^iθ) = ln(r) + iθ

In this case,

z = 1+i
z^2+1 = 1+2i = 2.2361 cis 63.4349°
√(z^2+1) = 1.4954 cis 31.7175° = 1.2720 + 0.7862i
z+√(z^2+1) = 2.2729 + 1.7862i = 2.8908 cis 38.1627° = 2.8908 cis 0.6312

ln(2.8908 cis 0.6312)
= ln(2.8908) + 0.6312i
= 1.0615 + 0.6312i

or j, as you engineers prefer...