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It is given that z1 = 3 cis(3π__4 ) and z2 = 2 cis(nπ__16) , n∈ ℤ+ .(a) In parts (a)(i) and (a)(ii), give your answers in the form r eiθ , r ≥ 0 , − π < θ ≤ π .(i) Find the value of z13

Question

It is given that z1 = 3 cis(3π__4 ) and z2 = 2 cis(nπ__16) , n∈ ℤ+ .(a) In parts (a)(i) and (a)(ii), give your answers in the form r eiθ , r ≥ 0 , − π < θ ≤ π .(i) Find the value of z13

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Solution

The given complex numbers are in polar form, where "cis" stands for "cosine plus i sine".

The complex number z1 = 3 cis(3π/4) can be written in exponential form as z1 = 3e^(i3π/4).

To find the value of z1^3, we use the property of exponents which states that (a^m)^n = a^(mn).

So, z1^3 = (3e^(i3π/4))^3 = 3^3 * e^(i(3*3π/4)) = 27e^(i9π/4).

However, we want our angle θ to be between -π and π. The angle 9π/4 is equivalent to π/4 when reduced to the interval from -π to π.

So, z1^3 = 27e^(iπ/4).

This problem has been solved

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