Need help with Complex Variables - Orthogonal Question?

Question

Let the function f(z) = u(x,y) + iv(x,y) be analytic in a domain D, and consider the families of leve curves u(x,y) = c_1 and v(x,y) = c_2, where c_1 and c_2 are arbitrary real constants. Prove that these families are orthogonal. More precisely, show that if z_0=(x_0, y_0) is a point D which is common to two particular curves u(x,y)=c_1 and v(x,y)=c2 and if f'(z_0) not equal to 0, the the lines tangent to those curves at (x_0, y_0) are perpendicular.

That is the long winded question. I know i have to show that the two gradient vectors are orthogonal, i.e., their dot product as vectors in the plane is zero, at any point of intersection between a level curve of u
and a level curve of v. I know how Cauchy-Riemann equations work, but not sure how to represent the functions as vectors since they are both constants - wouldn't du/dx=0 and so on ?

Answer 1

We know ∂u/∂x=∂v/∂y ann ∂u/∂y=-∂v/∂x

grad u=∂u/∂x i + ∂u/∂y i; grad v=∂v/∂x i + ∂v/∂y j

take the dot product

grad u ∙grad v=∂u/∂x*∂v/∂x+∂u/∂y*∂v/∂y

If these are perpendicular than this dot product will be zero

using Cauchy-Riemann, plug the product after the + sign with the C-R equalities and we have:
grad u ∙grad v = ∂u/∂x*∂v/∂x - ∂u/∂x*∂v/∂x =0 therefore perpendicular.

Answer 2

no you do not would desire to be conscious complicated numbers or complicated vector aspects. sin and cos are the two "vectors" that belong to the vector component of all non-provide up purposes over the era [a,b]. to show they're orthogonal, we strive this by using questioning cosmx and sinnx (the situation m and n are constants) in C[0,2pi] (the vector component of all non-provide up purposes over the era 0,2i (closed era). the indoors product defined in this vector section is the essential from 0 to 2pi of f(x)g(x)dx. to show they're orthogonal, you teach that the essential from 0 to 2pi of sin(nx)cos(mx)dx is comparable to 0 (it rather is with the aid of actuality the definition of two vectors being orthogonal is they're inner product is comparable to 0) yet you try this integration by using using trig identities