the possible number of equation Z=5
Answers
Answer:
It's a classic problem in many textbooks: Solve in complex numbers the equation
z5=z¯(1)
The solution is a as follows: We apply the modulus, and obtain that |z|=0, so z=0, which satisfies and second case |z|=1,so z¯=1z and the equation now has the form
z6=1(2)
whose solutions are the 6-th roots of unity. So final solution is
z=0, U6.
Ok, now my main concern: shouldn't we "check" the final solutions?
I mean, we applied the modulus, etc, found the possible value of modulus, then using the equation again and derived only a CONSEQUENCE of the original equation, so all the solutions of original equation are also solutions of z6=1 but not necessarily the other way around.
I mean, the checking is very simple, but what surprised me is that no solution I read does this check.
So I began to ask myself if there's something I don't see? Is that something obvious guaranteeing that equations (1) and (2) indeed have exactly the same solution set, regarding that, in the process of solution, equation (2) comes only a s a consequence of (1)? I mean, could I find another problem solved in the same manner, such that at the end it gives extraneous roots?
So my main question is about the equivalence of equations. What bothers me is that in the 4-5 textbooks I found this solutions, they check z=0, but don't bother to check U6.
Step-by-step explanation:
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Appearance .1. Verify the following for x = 2.y=-3 and 2-1
a. X* (y + z) XXy+XX Z
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b. x+(y+z) + (x + y) + (x+2)
a. Getting three heads b. getting only one head
c. getting two tails
d. getting no heads
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{TTT, TTH. THT. HTT, THH, HTH, HHT, TTT). Find the probability of the following:
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