x+yi=(a+bi) then x/a+y/b=e(a*a-b*b) then e=?
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Updated 7/21/2023 4:01:17 PM
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Answer: e = 1 / (a^2 - b^2)

Explanation: By comparing the real and imaginary parts, we find that x = a and y = 0. Substituting y = 0 into the given equation, we get 1 = e(a^2 - b^2), and solving for e gives us e = 1 / (a^2 - b^2).

Here's the detailed solution:
Given: x + yi = a + bi
Comparing the real parts:x = a
Comparing the imaginary parts: yi = bi
Now, we are given the equation: x/a + y/b = e(a^2 - b^2)
Substituting x = a and yi = bi: a / a + bi / b = e(a^2 - b^2)
Simplifying the left-hand side: 1 + i = e(a^2 - b^2)
Now, since we have a complex number on one side (1 + i) and the expression on the other side (e(a^2 - b^2)), they must be equal to each other, so their real and imaginary parts must be equal.
Comparing the real parts: 1 = e(a^2 - b^2) - -(1)
Comparing the imaginary parts: i = 0 (since there is no imaginary component on the right side) - -(2)
From equation (2), we can see that the imaginary part is 0, which means there is no "y" term in the original equation (x + yi = a + bi). This implies that y = 0.
Substituting y = 0 into equation (1): 1 = e(a^2 - b^2)
Now, solving for e: e = 1 / (a^2 - b^2)
So, e is equal to 1 divided by the difference of squares of "a" and "b".

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