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cosθ+i sinθ)4(sinθ+i cosθ)5cos𝜃+𝑖 sin𝜃4sin𝜃+𝑖 cos𝜃5   is equal to

Question

cosθ+i sinθ)4(sinθ+i cosθ)5cos𝜃+𝑖 sin𝜃4sin𝜃+𝑖 cos𝜃5   is equal to

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Solution

The given expression is in the form of Euler's formula, which states that e^(iθ) = cosθ + i sinθ.

Let's rewrite the given expression using Euler's formula:

(cosθ + i sinθ)^4 * (sinθ + i cosθ)^5

This can be rewritten as:

(e^(iθ))^4 * (e^(i(π/2 - θ)))^5

Simplify this to:

e^(4iθ) * e^(5i(π/2 - θ))

Using the property of exponents a^(m) * a^(n) = a^(m+n), we can add the exponents:

e^(4iθ + 5i(π/2 - θ))

This is the simplified form of the given expression. To find the exact value, you would need to know the value of θ.

This problem has been solved

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