Given the following expression \(\displaystyle{w}={j}{\cos{{\left[{\frac{{{1}}}{{{n}}}}{\arccos{{\left({\frac{{{j}}}{{{e}}}}\right)}}}+{\frac{{{m}\pi}}{{{n}}}}\right]}}}={j}{\cos{{\left({z}\right)}}}\)

Oxinailelpels3t14

Oxinailelpels3t14

Answered question

2022-03-24

Given the following expression
w=jcos[1narccos(je)+mπn]=jcos(z)

Answer & Explanation

Lesly Fernandez

Lesly Fernandez

Beginner2022-03-25Added 16 answers

Set z=x+iy, where x,y are real. Use the addition formula:
cos(x+iy)=cosxcosiysinxsiniy
Now you have
cosiy=coshy, siniy=isinhy
which is easy to prove in the way you suggested. So
icos(x+iy)=i(cosxcoshyisinxsinhy)
=icosxcoshy+sinxsinhy

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