Polarization of quadratic form yields sesquilinear form How does polarisation of any quadratic form $Q: V \to \mathbb{C}$ on a complex vector space $V$ yields a sesquilinear form? 
 A: The term quadratic form gets abused in this context.  A true quadratic form satisfies $Q(cv)=c^2Q(v)$, and determines a symmetric bilinear form via one of several equivalent polarization formulae, e.g. $B(u,v)=\frac{1}{4}(Q(u+v)-Q(u-v))$.
The term pseudo-quadratic form is applied to what you presumably mean, and satisfies $Q(cv)=\vert c\vert^2 Q(v)$.  A sesquilinear form has the property that $S(cv,cv)=\vert c\vert^2 S(v,v)$, so $S$ determines a pseudo-quadratic form $Q$ via $Q(v)=S(v,v)$.
In the quadratic case, the bilinear form is only uniquely determined if a constraint of symmetry is applied.  Similarly, in the pseudo-quadratic case, to achieve uniqueness generally we can apply the constraint that the form must be Hermitian, i.e. $H(u,v)=\overline{H(v,u)}$, but in the complex case we can actually recover the entire sesquilinear form with a modified polarization formula.
The polarization identity $S(u,v)=\frac{1}{4}(Q(u+v)-Q(u-v)+iQ(u+iv)-iQ(u-iv))$ determines the unique complex sesquilinear form satisfying $S(v,v)=Q(v)$ for a pseudo-quadratic form $Q$.
