Question

Cos x = 3i what is the value of x?

Comments

Cos x= 3i what is the value of x?

A relation we can use is,

e^(a + bi) = e^a * (cos(b) + i.sin(b))

Set b = π/2, then

e^(a + π/2.i) = e^a * (cos(π/2) + i.sin(π/2))

e^(a + π/2.i) = e^a * (0 + i.1) = e^a.i

e^(a + π/2.i) = e^a.i

We want cos(x) = 3i, therefore set e^a = 3, i.e. a = ln(3)

Then

e^(ln(3) + π/2.i) = 3i = cos(x)

giving,

x = arccos(e^(ln(3) + π/2.i))

If we want a general answer, then make b = π/2 + 2nπ, n = 1,2,3,... giving the general solution as,

x = arccos(e^(ln(3) + (π/2 + 2nπ).i)), n = 1,2,3,...

 

Answer 1

Cos x= 3i what is the value of x?

That last answer/comment wasn’t really very good.

All I was doing really was simply taking cos(x) = 3i and saying x = arcos(3i).

So, here’s another relation we can use.

cos(x) = (1/2)(e^(ix) + e^(-ix))

Let u = e^(ix) in the expression cos(x) = 3i, then

(1/2)(u + u^(-1)) = 31

u + u^(-1) 6i

u^2 – 6i.u + 1 = 0

u^2 – 6i.u + (-3i)^2 = (-3i)^2 -1

(u – 3i)^2 = -9 – 1 = -10

u = 3i +/- √(-10)

u = 3i +/- i.√(10)

Then, using e^(ix) = u,

e^(ix) = i(3 +/- √(10))

ix = ln{i(3 +/- √(10))}

x = -i.ln(i) – i.ln(3 +/- √(10))

x = -i.π/2.i – i.ln(3 +/- √(10))   using ln(i) = ln(1.e^(iπ/2)) = (π/2).i

x = π/2 – i.ln(3 +/- √(10))