Can we take conjugate on both side of the equation x-iy=root over a-ib/c-id; x-iy,a-ib,c-id are complex numbers. If yes explain it how?

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kshalike eNotes educator| Certified Educator

I may need some clarity on the equation you are asking about, so you may correct me if it does not look like this:

`x-iy=sqrt(a-ib)/(c-id)` . Also, I am not sure what you wish to solve for. However, we can talk about conjugates, to help you understand this and future problems. When we have a complex number in the denominator like c-id, we can multiply everything by the complex conjugate to turn the denominator into a non-zero real number. For a complex number c-id, the conjugate is c+id, where c (the real part) is not changed, and the sign of id (the imaginary part) is changed. If you multiply the denominator by a value, you must multiply the numerator by the same thing, as you know, so we are really just multiplying by a fraction that is equal to one.

`x-iy=sqrt(a-ib)/(c-id) * (c+id)/(c+id)` . When multiplying across the numerators and denominators, we can't combine the numerators because a-ib is in a square root and we can't use it to perform operations with something outside of a square root. But let's look at the denominators. We can multiply them, and the inside terms cancel each other. Squaring i gives us -1, so you can see that the conjugate is used to turn the fraction into a fraction with a real denominator instead of a complex one.

`(c-id)(c+id) = c^2 + c*id - id*c - i^2*d^2 = c^2 + d^2`

From `x-iy= sqrt(a-ib)/(c^2+d^2)`

What you would do would depend on the question, but as you can see, the conjugate is used with the complex denominator.

The conjugate would not be used on both side of the equation, but rather to simplify the fraction portion into something with a real denominator.