# 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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### 1 Answer

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. **