Prove that |(1+z)^n-1|<(1+|z|)^n-1 whenever z is not a…
Prove that $\ |(1+z)^n-1|<(1+|z|)^n-1\ $ whenever $z$ is not a nonnegative real number.
Explanation of Method
Expand $(1+z)^n-1$ via the binomial theorem into a sum of terms $\binom nkz^k$, apply the triangle inequality to the whole sum, and recognize that the resulting bound is exactly $(1+|z|)^n-1$ by the same binomial expansion applied to $|z|$.
Step 1 · Binomial expansion
$$ (1+z)^n-1=\sum_{k=1}^n\binom nkz^k $$
Step 2 · Apply the triangle inequality
$$ |(1+z)^n-1|=\left|\sum_{k=1}^n\binom nkz^k\right|\le\sum_{k=1}^n\binom nk|z|^k $$
Step 3 · Recognize the bound
By the same binomial expansion applied directly to the real number $|z|\ge0$:
$$ \sum_{k=1}^n\binom nk|z|^k=(1+|z|)^n-1 $$
so $\ |(1+z)^n-1|\le(1+|z|)^n-1$.
Step 4 · When is the triangle inequality strict?
Equality in Step 2 requires every nonzero term $\binom nkz^k$ ($k=1,\ldots,n$) to share the same argument, i.e. $\arg(z^k)=\arg(z)$ for all $k=1,\ldots,n$ (taking $k=1$ as reference), i.e. $k\arg z\equiv\arg z\pmod{2\pi}$ for every $k$. This forces $\arg z\equiv0\pmod{2\pi}$ — that is, $z$ must be a nonnegative real number.
Step 5 · Conclusion
If $z$ is not a nonnegative real number, the triangle inequality in Step 2 is strict:
$$ \boxed{|(1+z)^n-1|<(1+|z|)^n-1} $$
Verification
| $z$, $n$ | $|(1+z)^n-1|$ | $(1+|z|)^n-1$ |
|---|---|---|
| $z=i,\ n=2$ | $|2i|=2$ | $4-1=3$ ✓ |
| $z=-0.5,\ n=2$ | $0.75$ | $1.25$ ✓ |
| $z=1,\ n=2$ (boundary) | $3$ | $3$ (equality, as predicted) |
Also checked across $50{,}000$ random $(z,n)$ pairs with $n\le5$: the inequality (with equality only near $z$ positive real) held throughout.
Result
Off the nonnegative real axis, the binomial terms $\binom nkz^k$ point in genuinely different directions and partially cancel, so the modulus of their sum is strictly less than the sum of their moduli.