Cauchy-Riemann equation for f(z) = 2x + ixy²

[RJLiberator]

I am currently learning how to work with Cauchy-Riemann equations.

The equation is f(z) = 2x+ixy^2.

My question: is u(x,y) = 2x or just x?
At this link: http://www.math.mun.ca/~mkondra/coan/as3a.pdf in letter e) they say u(x,y) is equal to x. But I don't understand how that is possible.

Is that a typo or am I missing something critically important?

Thank you.

 

[WWGD]

I am 99.999997% that this should be $u(x,y)=2x$ . It is a typo; Given f(x,y)=u(x,y)+iv(x,y), u(x,y) is the Real part of f(x,y)..

 

[RJLiberator]

Excellent. Thank you for that confirmation.

 

[HallsofIvy]

I presume that you text has already defined x and y as the real and imaginary parts of z, z= x+ iy, so that x and y are real numbers themselves.

 

[RJLiberator]

Yes, that is correct indeed.