This is to complement the inequality $$|(1+it)^n-1|\ge b_1(n,t):=|e^{int}-1|,\tag{10}\label{10}$$ proved by Terry Tao for real $t$ and natural $n$, by the following inequality: $$|(1+it)^n-1|\ge b_2(n,t):=(1+t^2)^{n/2}-1\ge b_3(n,t):=nt^2/2 \tag{20}\label{20}$$ for real $t$ and real $n\ge0$.
For any fixed real $t\ne0$, the lower bound $b_2(n,t)$ on $|(1+it)^n-1|$ grows exponentially in $n$, whereas the lower bound $b_1(n,t)$ remains bounded by $2$. More generally, the lower bound $b_2(n,t)$ (and even the lower bound $b_3(n,t)$) will be better than $b_1(n,t)$ if $nt^2$ is large enough (say if $nt^2>4$).
To prove \eqref{2}, just note that $$|(1+it)^n-1|^2=1 - 2 c (1 + t^2)^{n/2} + (1 + t^2)^n\ge b_2(n,t)^2,$$ where $c:=\cos(n\arctan t)\le1$.
It is now also seen that, for any real $T\ge\tan\dfrac{2\pi}n$, there is some $t\in(0,T]$ such that the lower bound $b_2(n,t)$ on $|(1+it)^n-1|$ is exact, in the sense that $|(1+it)^n-1|=b_2(n,t)$. Actually, the lower bound $b_2(n,t)$ on $|(1+it)^n-1|$ is exact for all $t=\tan\dfrac{2\pi k}n$, where $k=0,1,\dots$.
Here are the graphs $\Big\{\Big(t,\dfrac{|(1+it)^n-1|}{b_1(n,t)}\Big)\colon0<t<0.6,b_1(n,t)\ge b_2(n,t)\}$ (gold) and $\Big\{\Big(t,\dfrac{|(1+it)^n-1|}{b_2(n,t)}\Big)\colon0<t<0.6,b_2(n,t)\ge b_1(n,t)\}$ (blue) for $n=100$:
- gold $$ \Big\{\Big(t,\dfrac{|(1+it)^n-1|}{b_1(n,t)}\Big)\colon0<t<0.6,b_1(n,t)\ge b_2(n,t)\Big\} $$ and
- blue $$ \Big\{\Big(t,\dfrac{|(1+it)^n-1|}{b_2(n,t)}\Big)\colon0<t<0.6,b_2(n,t)\ge b_1(n,t)\Big\} $$
