Q.Find $(2-i)(3+4i)$ and express the result in the form $a+ib$.

A complex number is an expression $z=a+ib$, where $a,b$ are real numbers and $i$ is the imaginary unit defined by $i^2=-1$. The number $a=\mathrm{Re}(z)$ is its real part and $b=\mathrm{Im}(z)$ is its imaginary part; two complex numbers are equal exactly when both parts match separately. Complex numbers were introduced precisely because equations like $x^2+1=0$ have no real solution — the real numbers cannot supply a square root of a negative number, but $\mathbb{C}$ can. Addition and subtraction combine real parts with real parts and imaginary parts with imaginary parts: $(a+ib)\pm(c+id)=(a\pm c)+i(b\pm d)$. Multiplication expands like ordinary binomials with the substitution $i^2=-1$: $(a+ib)(c+id)=(ac-bd)+i(ad+bc)$. The conjugate $\bar z=a-ib$ satisfies $z\bar z=a^2+b^2$, and this identity is the key to finding a reciprocal: $z^{-1}=\bar z/|z|^2$, which in turn gives division of one complex number by another by multiplying through by the conjugate of the denominator.