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What is the slope of the image of the st. line $y=mx+c$ under the mapping $f(z)=az+b,a,b\in\mathbb C?$

I think I don't need the constant $b$ for translation don't rotates any shape. Letting $A$ is the angle the line makes with the postive real axis I have found that the image line makes the angle $A+\arg a$ with the positive $x$-axis as follows:

enter image description here

Thus the slope of the new line is $\tan (A+\arg a)=\tan(\tan^{-1}m+\arg a).$ Where did I go wrong?

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When we talk about the image of a complex valued function, we are talking about the image in $u,v$ plane.
$$f(z) = a(x+iy)+b = a(x+i(mx+c))+b = (ax+b)+a(mx+c)i = u(x)+i\cdot v(x)$$ $$u(x) = ax+b,\;\;v(x) = a(mx+c)$$ So the image of $f$ will be $$\{(ax+b,a(mx+c))\;|\;x\in\mathbb{R}\}$$

Notice while $x$ takes all value in $\mathbb{R}$, so will $u$. We have $v = a(mx+c) = mu+(ac-mb)$.

The image then can be rewritten as $$\{(u,mu+(ac-mb)\;|\;u\in\mathbb{R}\}$$

That is, in the $uv$ plane, it looks like the graph of linear affine $v = mu+(ac-mb)$, a line with slope $m$ and intercept $(ac-mb)$. Therefore the slope is still $m$.

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