$\int_{0}^{1+2i}z^{2} dz$, along the curve $2x^{2}=y$

Solution:

$f(z)=z^{2}$

Let $z=x+iy$ -------------(1)

$f(z)=(x+iy)^{2}=x^{2}+2ixy+i^{2}y^{2}=x^{2}+2ixy-y^{2}$

But $y=2x^{2}$

$f(z)=x^{2}+2ix.2x^{2}-(2x^{2})^{2}=x^{2}+4ix^{3}-4x^{4}$ ----------------(2)

Differentiating (1) w.r.t. z,

$dz=dx+i \ dy$ ----------------(3)

But $y=2x^{2}$, therefore $dy=4x \ dx$

Put dy in (3),

$dz=dx+ i4x \ dx = (1+i4x)dx$ ----------(4)

$\therefore \int_{c}f(z) \ dz= \int_{0}^{1} (x^{2}+4ix^{3}-4x^{4})(1+i4x)dx = \int_{0}^{1} [x^{2}+4ix^{3}-4x^{4}+4ix^{3}-16x^{4} - 16ix^{5}] \ dx$

$= \int_{0}^{1} [(x^{2}-4x^{4}-16x^{4})-i(-4x^{3}-4x^{3} + 16x^{5})] \ dx$

$= \int_{0}^{1} [(x^{2}-20x^{4})-i(-8x^{3} + 16x^{5})] \ dx$

$=\left [ \left[ \cfrac{x^{3}}{3} - \cfrac{20x^{5}}{5} \right] - i \left[ -\cfrac{8x^{4}}{4} + \cfrac{16x^{4}}{6} \right ] \right]_{0}^{1}$

$=\left [ \left[ \cfrac{1}{3} - \cfrac{20}{5} \right] - i \left[ -\cfrac{8}{4} + \cfrac{16}{6} \right ] \right] - \left [ \left[ 0-0 \right] - i \left[ 0-0 \right ] \right]$

$=\left [ \left[ \cfrac{1}{3} - 4 \right] - i \left[ -2 + \cfrac{8}{3} \right ] \right]$

$=\left [ \cfrac{11}{3} \right] - i \left[ \cfrac{2}{3} \right ] = - \cfrac{11}{3} - \cfrac{2i}{3}= \cfrac{-11-2i}{3} $