At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Sorry I MEAN, show that
\[|\cos z|^{2} = \cos^{2} x + \sinh^{2}y\]

Take LHS first
(cosz)^2
replace z with x +iy
cos(x+iy) = cos(x)cos(iy) - sin(x)sin(iy)

as cos(iy) = coshy and sin(iy) =i sinhy
therefore
cosz = cos(x)cos(hy)-i sin(x)sin(hy)

@gerryliyana after squaring and applying (a-b)^2 formula , the equation become difficult to solve

i've tried.., but didn't match,

how did u approach the problem?

using hint above..., i tried, one by one.., hbu?? it's make sense ?

how can u apply above hint in RHS?

what the next ??

@UnkleRhaukus help?

have idea ???

but it's not the answer

as you see, the LHS and the RHS differ by \(1\over2\).

yes., i see, did you know why ???