Is this the correct answer for cos(z) = sqrt(2)?

Question

This is the question and my working out, can anyone please tell me if its correct? Thanks.

cos(z) = sqrt(2)

so

z = cos^(-1)(sqrt(2))

now cos^(-1)(z) = -i * log(z+i(1-z^2)^(1/2))

and z = sqrt(2)

so cos^(-1)(z) = i log(sqrt(2) + i(1-2)^(1/2))
= i * log(sqrt(2) + i * i)
= i * log(sqrt(2) -1)

Is that it?

also, heaps of you guys change log into ln, when is it okay to do that - for example in my answer? Can I just change it?


Thanks, Laura

Hi There, no it's not meant to be that easy, it's def meant to be =sqrt(2) its in the text book in the chapter on inverse trig functions

Answer 1

I only remember one thing:
e^(ix) = cos(x) + i*sin(x)
which means
e^(-ix) = cos(x) - i*sin(x)
so
cos(x) = (e^(ix)+e^(-ix))/2
You might have already known that, but I always have to derive it.

Your problem says
cos(z) = √2 = v
so let q = e^(iz)
Then (q+1/q)/2 = v so
q² + 1 = 2qv or
q² - 2qv + 1 = 0
Thus q = (2v +/- √(4v²-4))/2 = v +/- √(v²-1) = √2 +/- 1

e^(iz) = v +/- √(v²-1) so taking logs =>
iz = ln (v +/- √(v²-1)) so multiplying by -i =>
z = -i * ln (v +/- √(v²-1)) = -i * ln (√2 +/- 1)

-> ln <- means natural log = Naperian log = log base e.
If you write log(x), it means log of x where the base is somewhat arbitrary, typically 10 or 2, but it is up to the author to specify.

Answer 2

|cos z| should be <= 1 so feel that question is incorrect because 1 < √2 ?
Perhaps question should be:-
cos z = 1 / √2
z = π/4 , 7π/4