Is there a complex number that is equal to the square of its conjugate?

Answer 1

Oh, dude, you're hitting me with some math vibes! So, like, a complex number is in the form a + bi, right? The conjugate of that is a - bi. If you square a + bi, you get (a^2 - b^2) + 2abi. The conjugate of that is (a^2 - b^2) - 2abi. So, for a complex number to be equal to the square of its conjugate, you'd need (a^2 - b^2) + 2abi = (a^2 - b^2) - 2abi, which means b has to be 0. So, the complex number would be a real number.

Answer 2

Yes, a complex number that is equal to the square of its conjugate must satisfy the equation z = (z*)^2, where z is a complex number and z* is its conjugate. By expanding z = (a + bi)^2 and (z*)^2 = (a - bi)^2, where a and b are real numbers, we can equate the two expressions to find a solution. Solving the equation yields z = 0 or z = 1, where z = 0 is the trivial solution and z = 1 is a non-trivial solution.

Answer 3

Oh, isn't that a lovely question! When you square a complex number and find it equal to its conjugate, it means the number is on the real axis. So, the complex number you're looking for is a real number. Just like a happy little tree standing tall and proud!

Answer 4

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.