$(y^2+2iy)^{m-\frac{1}{2}}=y^{m-\frac{1}{2}}(2i)^{m-\frac{1}{2}}+O(y^{m+\frac{1}{2}})(0\leq y \leq 1)?$ This question comes from stein's book Introduction to fourier analysis on euclidean space,page 159.
when $m > 1/2$,
$$(y^2+2iy)^{m-\frac{1}{2}}=y^{m-\frac{1}{2}}(2i)^{m-\frac{1}{2}}+O(y^{m+\frac{1}{2}})(0\leq y \leq 1)$$
$$(y^2+2iy)^{m-\frac{1}{2}}=y^{m-\frac{1}{2}}(2i)^{m-\frac{1}{2}}+O(y^{2m-1})(1\leq y \leq \infty)$$
why $(y^2+2iy)^{m-\frac{1}{2}}$ can be written in this way? It is not obviously at all.

The first equality comes from the fact $\forall \alpha>0, (1+x)=_01+O(x)$ where $=_0$ means that the equality is true for $x$ in a neighborhood of $0$. Then we have:
$$(y^2+2iy)^{m-\frac{1}{2}}=(2iy)^{m-\frac{1}{2}}\left((\frac{1}{2i})^{m-\frac{1}{2}}y+1\right)=_0 (2iy)^{m-\frac{1}{2}}(1+O(y))\\=_0y^{m-\frac{1}{2}}(2i)^{m-\frac{1}{2}}+O(y^{m+\frac{1}{2}}).$$
The second equality is a little weird, I find this result:
$$(y^2+2iy)^{m-\frac{1}{2}}=(y^2)^{m-\frac{1}{2}}(1+\frac{2i}{y})^{m-\frac{1}{2}}=_{+\infty}(y^2)^{m-\frac{1}{2}}(1+O(\frac{1}{y}))\\=_{+\infty}y^{2m-1}+O(y^{2m-2}).$$
If there is a correction, it is welcome.
