Complex conjugate

The complex conjugate of $z=x+iy$ is

$\bar{z}=x-iy,$

obtained by flipping the sign of the imaginary part. Geometrically, $\bar{z}$ is the reflection of $z$ across the real axis of the complex plane.

The crucial identity

$z\bar{z}=(x+iy)(x-iy)=x^2-(iy{)}^2=x^2+y^2=|z{|}^2.$

The product of $z$ with its conjugate is real and non-negative, equal to the squared modulus. This is what makes the conjugate useful: multiplying numerator and denominator of $z_1/z_2$ by ${\bar{z}}_2$ clears $i$ out of the denominator.

$\frac{z_1}{z_2}=\frac{z_1{\bar{z}}_2}{z_2{\bar{z}}_2}=\frac{z_1{\bar{z}}_2}{|z_2{|}^2}.$

Picking off real and imaginary parts

$z+\bar{z}=2\mathrm{Re}(z),z-\bar{z}=2i\mathrm{Im}(z).$

So $\mathrm{Re}(z)=(z+\bar{z})/2$ and $\mathrm{Im}(z)=(z-\bar{z})/(2i)$. Specialized to $z=e^{i\theta }$, these become the Euler formulas for cosine and sine:

$\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2},\sin \theta =\frac{e^{i\theta }-e^{-i\theta }}{2i}.$

See Euler’s formula.

Conjugate respects arithmetic

$\overline{z_1+z_2}={\bar{z}}_1+{\bar{z}}_2,\overline{z_1{z}_2}={\bar{z}}_1\cdot {\bar{z}}_2,\overline{1/z}=1/\bar{z}.$

In polar form: if $z=r{e}^{i\theta }$, then $\bar{z}=r{e}^{-i\theta }$ — same magnitude, opposite argument.

A function, not just an operation

Viewed as a function $f(z)=\bar{z}$, conjugation is continuous everywhere on $\mathbb{C}$ but complex differentiable nowhere. Along the real axis the difference quotient $\overline{\Delta z}/\Delta z$ equals $1$; along the imaginary axis it equals $-1$. Different directional limits, so $f$ fails the complex derivative test at every point. Geometrically, conjugation is a reflection — orientation-reversing — while complex differentiability requires the function to behave locally like rotation-and-scaling, which preserves orientation.

This is one of the cleanest examples of a function that is “smooth” in every real-variable sense yet pathological from the complex-analytic viewpoint. It motivates the Cauchy-Riemann equations.

In phasor language

If $V$ is the phasor of $v(t)=A\cos (\omega t+\varphi )$, then $\bar{V}$ is the phasor of $A\cos (\omega t-\varphi )$ — same amplitude, mirrored phase. Summing $V$ with $\bar{V}$ gives $2A\cos \varphi$, real, which corresponds to two phasors symmetric about the real axis combining to a pure cosine without phase shift. See Phasor.