z1=2(cos(pi/5)+isin(pi/5)) and z2=8(cos(7pi/6)+sin(7pi/6)) How do I calculate for z1z2, z2 , z1 using the equation?

Expert Answers

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We can represent compex numbers in polar form. A complex number z=a+bi is represented as `z=r"cis" theta=r(costheta+isintheta)` where r=|z| (|z| is the modulus or absolute value of z; it is the length of the vector in polar coordinates.)

(1) To find the product of two complex numbers in polar form we multiply the absolute values and add the polar angles. The formula:

If `z_1=r"cis"alpha,z_2=s"cis"beta` then `z_1z_2=rs"cis"(alpha+beta)` or `z_1z_2=rs(cos(alpha+beta)+isin(alpha+beta))`

So given `z_1=2(cos(pi/5)+isin(pi/5)),z_2=8(cos(7pi)/6+isin(7pi)/6)` we can write `z_1=2"cis"pi/5,z_2=8"cis"(7pi)/6`

Then `z_1z_2=(2*8)("cis"(pi/5+(7pi)/6))=16"cis"(41pi)/30`

Alternatively, you could multiply as normal:


`=(2*8)(cos(pi/5)cos((7pi)/6)+icos(pi/5)sin((7pi)/6)+isin(pi/5)cos((7pi)/6)+i^2sin(pi/5)sin((7pi)/6))` Rearranging terms we get:

`=16(cos(pi/5)cos((7pi)/6)-sin(pi/5)sin((7pi)/6)+i[sin(pi/5)sin((7pi)/6)+cos(pi/5)cos((7pi)/6)])` `=16(cos(pi/5+(7pi)/6)+isin(pi/5+(7pi)/6))` Using the angle sum formulas



** The other power use for this notation is finding powers of complex numbers. If `z=r"cis"theta` then `z^n=r^n"cis"(n theta)` ; just raise the modulus to the nth power, and multiply the angle by n.

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