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Consider the function $f(z)=z^2$. Prove that level curves of $Re(f(z))$ and $Im(f(z))$ at $z=1+2i$ are orthogonal to each other.

I am not sure how to apply level curves or contour lines for complex variables. As far as real variables go, I am aware that for a function like $f(x,y) = \sqrt{x^2+y^2}$, the level curves are the circles centered at $(0,0)$. For this question would it be sufficient for me to take the gradients of the real components and the imaginary components and show that its dot product equals to $0$? But if I do that, how am I showing that the level curves are orthogonal to each other?

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The map $\Bbb C\to\Bbb R^2$ that takes real/imaginary parts to first/second components will preserve the geometry of figures in the complex plane. Like: curves, intersections and orthogonality. –  anon May 16 '12 at 22:01

1 Answer 1

up vote 3 down vote accepted

How much do you know about holomorphic or analytic functions? Write $z = x+iy$ and split $f(z)$ into its real and imaginary part: $$f(z) = f(x+iy) = (x+iy)^2 = (x^2 - y^2) + 2ixy,$$ so $u(x,y) = x^2-y^2$ is the real part of $f$ and $v(x,y) = 2xy$ is the imaginary part. Compute the gradient of $u$ and $v$ and check that they are orthgonal at $(x,y) = (1,2)$.

Once you've done this, you will probably realize that there is nothing too special about the function, and nothing special at all about the specific point. The key observation here is Cauchy-Riemann's equations ($u'_x = v'_y$ and $u'_y = -v'_x$) and these are satisfied by every analytic function.

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