If |z| < dfrac12 , show that |(1+i)z^3+iz| < dfrac34
If $|z| < \dfrac12$, show that $|(1+i)z^3+iz| < \dfrac34$.
Explanation of Method
Bound the expression using the triangle inequality and multiplicativity of the modulus, reducing everything to a real-valued function of $r=|z|$. Since that function turns out to be strictly increasing, its supremum over $0\le r<\frac12$ is its value at the (excluded) endpoint $r=\frac12$ — checking that this endpoint value is already below $\frac34$ finishes the proof.
Step 1 · Apply the triangle inequality
$$ |(1+i)z^3+iz| \le |(1+i)z^3| + |iz| $$
Step 2 · Simplify each term
Using multiplicativity of the modulus and $|1+i|=\sqrt{1^2+1^2}=\sqrt2$, $|i|=1$:
$$ |(1+i)z^3| = \sqrt2\,|z|^3, \qquad |iz| = |z| $$
so:
$$ |(1+i)z^3+iz| \le \sqrt2\,|z|^3+|z| = |z|\left(\sqrt2\,|z|^2+1\right) $$
Step 3 · Reduce to a function of $r=|z|$
Let $r=|z|\in[0,\tfrac12)$ and define:
$$ f(r) = \sqrt2\,r^3+r $$
We must show $f(r) < \dfrac34$ for every $r<\dfrac12$.
Step 4 · Show $f$ is increasing
$$ f'(r) = 3\sqrt2\,r^2+1 > 0 \quad \text{for all } r\ge0 $$
so $f$ is strictly increasing on $[0,\infty)$. Consequently, for $r<\frac12$:
$$ f(r) < f\!\left(\frac12\right) $$
Step 5 · Evaluate at the boundary $r=\frac12$
$$ f\!\left(\frac12\right) = \sqrt2\cdot\frac18+\frac12 = \frac{\sqrt2}{8}+\frac48 = \frac{\sqrt2+4}{8} $$
Step 6 · Compare to $\frac34$
Check $\dfrac{\sqrt2+4}{8} < \dfrac34 = \dfrac68$, i.e. $\sqrt2+4<6$, i.e. $\sqrt2<2$ — true, since $\sqrt2\approx1.414$.
So:
$$ f\!\left(\frac12\right) = \frac{\sqrt2+4}{8} \approx 0.677 < 0.75 = \frac34 $$
Step 7 · Combine
For $|z|<\frac12$, chaining Steps 2, 4, and 6:
$$ |(1+i)z^3+iz| \le f(|z|) < f\!\left(\frac12\right) < \frac34 $$
$$ \boxed{|(1+i)z^3+iz| < \frac34} $$
$\blacksquare$
Step 8 · Numeric spot check
For $z=0.4$ (real, satisfying $|z|<\tfrac12$):
$$ f(0.4) = \sqrt2(0.4)^3+0.4 \approx 1.414(0.064)+0.4 \approx 0.0905+0.4 = 0.4905 < 0.75 \quad \checkmark $$
consistent with the bound. Note also that the actual supremum of the right side, $\dfrac{\sqrt2+4}{8}\approx0.677$, is comfortably below $\dfrac34=0.75$ — so the stated bound of $\dfrac34$, while true, is not the tightest possible one; the sharp bound (via this triangle-inequality argument) is $\dfrac{\sqrt2+4}{8}$.
Result
For $|z|<\dfrac12$:
proved via the triangle inequality, monotonicity of $f(r)=\sqrt2 r^3+r$, and the numeric check $\sqrt2<2$; the argument in fact yields the slightly tighter bound $\dfrac{\sqrt2+4}{8}\approx0.677$.