To write the complex number z = 2(cos(5Ï€/6) + sin(5Ï€/6)) in rectangular form, we need to convert it from polar form to rectangular form.
First, let's recall Euler's formula:
e^ix = cos(x) + i*sin(x)
Now, let's rewrite the given expression in polar form:
z = 2(cos(5Ï€/6) + sin(5Ï€/6))
We can convert cos(5Ï€/6) and sin(5Ï€/6) to their exponential forms:
cos(5Ï€/6) = Re[e^(i * (5Ï€/6))]
sin(5Ï€/6) = Im[e^(i * (5Ï€/6))]
Now, let's rewrite the expression with the exponential forms:
z = 2[ Re[e^(i * (5Ï€/6))] + Im[e^(i * (5Ï€/6))] ]
Using Euler's formula, we get:
z = 2[ Re[e^(i * (5Ï€/6))] + Im[e^(i * (5Ï€/6))] ]
= 2 * e^(i * (5Ï€/6))
Next, let's simplify the exponential form:
e^(i * (5π/6)) = e^(i * 5π) * e^(i * π/6)
= e^(i * π) * e^(i * π/6)
= -1 * e^(i * π/6)
= -e^(i * π/6)
Therefore, z = 2 * (-e^(i * π/6))
To write it in rectangular form, we can substitute e^(i * π/6) with its rectangular form expression. The rectangular form of a complex number a + bi is simply (a, b).
Using the identity e^(ix) = cos(x) + i*sin(x), we can determine the rectangular form of -e^(i * π/6) as:
Re[-e^(i * π/6)] = -cos(π/6)
Im[-e^(i * π/6)] = -sin(π/6)
Therefore, z = 2 * (-cos(Ï€/6) - sin(Ï€/6))
Simplifying further, we get:
z = -2 * (cos(Ï€/6) + sin(Ï€/6))
Thus, the rectangular form of the complex number z is:
z = -2 * (cos(Ï€/6) + sin(Ï€/6))