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Is there a complex number that is equal to the square of its conjugate?
Wiki User
∙ 12y ago
If the number is (a + bi) then the conjugate is (a - bi)
so set (a + bi) = (a - bi)² = a² - 2abi - b².
This can be split into to separate equations, because the real part on the left must equal the real part on the right, and the imaginary part on the left must equal the imaginary on the right.
a = a² - b² ; and b = -2ab.
Use the 2nd equation to solve for a, by dividing both sides by b: 1 = -2a ---> a = -1/2.
Plug this into the first equation, and solve for b: -1/2 = (-1/2)² - b² --- b² = 3/4.
So b = (±√3)/2, So the number -1/2 + i(√3)/2, and its conjugate -1/2 - i(√3)/2, will solve the conditions.
Wiki User
∙ 12y ago
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