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hamberlona
9 months ago

Yes, it seems to be right. The confusing: There are not all solutions with all k equal as they would be if k*2*PI were added outside the break in the exponent. Nevertheless, all meet the condition: if you potentize them with 4 the z^4 comes out in the penultimate line. The amount of solution looks like an inclined +, and if you take the high 4 each right angle spreads to a full rotation.

Halswirbelstrom
9 months ago

LG H.