Question

line integral

Answer 1

\[\int\limits_{0}^{1+2i} |z|^2 dz\]

Answer 2

\[\int\limits_{0}^{1+2i} (x^2-y^2)(dx+idy)\]

Answer 3

@ikram002p @zepdrix

Answer 4

Parameterize the contour by \(\gamma(t)=\langle t,2t \rangle\) for \(0\le t\le1\), where \(x(t)=t\) and \(y(t)=2t\). With \(z(t)=x(t)+iy(t)\), you have
\[\int_\gamma |z|^2\,\mathrm{d}z=\int_\gamma |x+iy|^2\,\mathrm{d}(x+iy)=\int_0^1(t^2-4t^2)(1+2i)\,\mathrm{d}t\]

Answer 5

I get -1-2i, but the answer is 5/3(1+2i)

Answer 6

@SithsAndGiggles

Answer 7

Sorry, I carried on from your work. It should actually be \(|z|^2=x^2\color{red}+y^2\), since \(\displaystyle |z|=\sqrt{\mathrm{Re}(z)^2+\mathrm{Im}(z)^2}=\sqrt{x^2+y^2}\).

Answer 8

I see, but using the z(t) =(1-t)p +qt, where p is starting point and q is ending point, should give me same answer right?

Answer 9

z(t)= (1-t)(0)+(1+2i)t
z(t)= t+2ti
but when I square I get
t^2-4t^2

Answer 10

Oh, I get it. Is it because by definition of img is real and we don't include i?

Answer 11

Exactly.