# Why is L(z)=Az+B a linear transformation in Complex Analysis?

Recently, I am learning complex analysis using " Complex Analysis for mathematics and engineering" by John H. Mathews and Russell W. Howell. In Chapter 2, it says that "...because a linear transformation can be considered to be a composition of a rotation, a magnification, and a translation", where in the text a linear transformation is defined as w=L(z)=Az+B. This is really confusing, because in linear algebra, this doesn't make sense. So, is the definition of linear transformation in complex number different from that of linear algebra? Thank you!

• This comes from the definition of a linear function. $+b$ is the translation. $a$ is a complex number and every complex number can be written in polar form which can be realized as a rescaling and rotation. Jun 16, 2018 at 1:43
• Well, I think that I didn't express my question nicely! The question is edited. I am confused why the way of defining linear transformation in complex number can be so explicitly written in L(z)=Az+B, where A and B are complex numbers.
– Ivy
Jun 16, 2018 at 1:52
• @CameronWilliams I understood! Is it because it is a one-dimensional linear transformation over the field C? Thank you!
– Ivy
Jun 16, 2018 at 1:54
• I'm not sure what you mean by linear transformations not having such an explicit definitions: Linear maps from $\mathbb{R}$ to itself are given by $f(x)=ax$ for $a$ a real number. As mentioned, colloquially we say a funciton is linear if it is really any affine function, i.e. we allow for translations (basically we just want the graph to be a line, $f(x)=ax+b$) Jun 16, 2018 at 2:13
• Affine is a fine word. Jun 16, 2018 at 2:49

There is a very sad confusion that arises naturally as a result of poor choice of terminology in English mathematics, which is calling functions of the form $f (x)=ax+b$ linear functions - $x$ be a real or complex number, it doesn't matter.
The reason it is unfortunate is that these functions are not necessarily linear transformations, since $f (0)=b$ is not necessarily $0$.