Conjugate, Multiplicative Inverse, and Square Roots of a Complex Number

The conjugate of $z=a+ib$, written $\bar z$, is obtained by flipping the sign of the imaginary part: $\bar z = a - ib$. Geometrically (as the next sections will make precise) this is a mirror reflection across the real axis, but algebraically its power is that it turns complex expressions into real ones — that is exactly the trick used for division above, since $z\bar z = (a+ib)(a-ib) = a^2+b^2$ is always a non-negative real number.

Conjugation interacts predictably with the other operations, and these rules are used constantly in simplifying expressions:

Each follows by writing $\alpha=a+ib$, $\beta=c+id$ and expanding both sides directly — conjugation of a sum/product/quotient is the sum/product/quotient of the conjugates.

Using $z\bar z = a^2+b^2$, the multiplicative inverse of any nonzero $z=a+ib$ can be written compactly as $z^{-1} = \dfrac{\bar z}{z\bar z} = \dfrac{a-ib}{a^2+b^2}$ — the same formula reached in Section 1.2 by a different route, now expressed using the conjugate.

Square roots of a complex number. Given $z=a+ib$, we look for $w=x+iy$ with $w^2=z$, i.e. $(x+iy)^2 = a+ib$. Expanding, $x^2-y^2 = a$ and $2xy=b$. Solving this pair of real equations (using $x^2+y^2 = |a+ib| = \sqrt{a^2+b^2}$, from the modulus, covered next) gives the standard square-root formulas:

with the sign of $y$ chosen to match the sign of $b$ (so that $2xy=b$ actually holds) — a positive $b$ needs $x,y$ of the same sign, a negative $b$ needs opposite signs. Every nonzero complex number has exactly two square roots, negatives of each other, matching the real-number pattern where $\sqrt{4}=\pm2$. …