For k=1,2,…9, if we define zk=cos(3kπ/10)+isin(2kπ/10), then is it possible that z1×z=zk has no Solution z?

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answerhappygod
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For k=1,2,…9, if we define zk=cos(3kπ/10)+isin(2kπ/10), then is it possible that z1×z=zk has no Solution z?

Post by answerhappygod »

a) True
b) False
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