Show that the point z = 1+i-α-iβ, for 0<α<1, 0<β<1, 0<α+β<1…
Show that the point $$ z = 1+i-\alpha-i\beta, \qquad \text{for} \qquad 0<\alpha<1,\ \ 0<\beta<1,\ \ 0<\alpha+\beta<1 $$ is located in the interior of the triangle with vertices $1,\ 1+i,\ i$.
Explanation of Method
A point lies strictly inside a triangle exactly when it can be written as a convex combination of the three vertices, $z = \lambda_1 A + \lambda_2 B + \lambda_3 C$, with all $\lambda_i > 0$ and $\lambda_1+\lambda_2+\lambda_3=1$. The strategy is to write $z$ in that form and check the coefficients are positive under the given constraints on $\alpha,\beta$. As a cross-check, the same conclusion follows from testing $z$ against the half-plane defined by each edge of the triangle.
Step 1 · Identify the vertices
$$ A = 1 = (1,0), \qquad B = 1+i = (1,1), \qquad C = i = (0,1) $$
This is a right triangle with legs along $x=1$ (from $A$ to $B$) and $y=1$ (from $B$ to $C$), and hypotenuse from $C$ to $A$ along the line $x+y=1$.
Step 2 · Write $z$ in Cartesian form
$$ z = 1+i-\alpha-i\beta = (1-\alpha) + i(1-\beta) $$
so $z = (x,y)$ with $x = 1-\alpha$, $y = 1-\beta$.
Step 3 · Write $z$ as a convex combination of the vertices
Try coefficients $\lambda_A = \beta$, $\lambda_C = \alpha$, $\lambda_B = 1-\alpha-\beta$ for vertices $A=1$, $C=i$, $B=1+i$ respectively, and check they reproduce $z$:
$$ \beta\cdot 1 + \alpha\cdot i + (1-\alpha-\beta)(1+i) = \big[\beta + (1-\alpha-\beta)\big] + i\big[\alpha + (1-\alpha-\beta)\big] $$
$$ = (1-\alpha) + i(1-\beta) = z \quad \checkmark $$
So indeed:
$$ \boxed{z = \beta\cdot 1 \;+\; \alpha\cdot i \;+\; (1-\alpha-\beta)\cdot(1+i)} $$
The three coefficients sum to $\beta+\alpha+(1-\alpha-\beta)=1$, and each is strictly positive: $\beta>0$, $\alpha>0$, and $1-\alpha-\beta>0$ is exactly the given condition $\alpha+\beta<1$. Since $z$ is a strict convex combination of the three vertices, it lies in the interior of the triangle.
Step 4 · Cross-check with the three edge inequalities
As an independent check, a point $(x,y)$ lies inside this triangle exactly when it satisfies all three strict inequalities: $x<1$ (left of edge $AB$: $x=1$), $y<1$ (below edge $BC$: $y=1$), and $x+y>1$ (above edge $CA$: $x+y=1$).
Edge $AB$ ($x=1$): from $0<\alpha<1$, $x = 1-\alpha < 1$. ✓
Edge $BC$ ($y=1$): from $0<\beta<1$, $y = 1-\beta < 1$. ✓
Edge $CA$ ($x+y=1$): $x+y = (1-\alpha)+(1-\beta) = 2-(\alpha+\beta) > 2-1 = 1$, using $\alpha+\beta<1$. ✓
Also $x = 1-\alpha > 0$ and $y = 1-\beta > 0$ follow from $\alpha<1$, $\beta<1$, so the point sits in the first quadrant as required. All three edge conditions hold, confirming $z$ is in the interior.
The triangle with vertices $1$, $1+i$, $i$, with an example point $z=0.7+0.6i$ (from $\alpha=0.3,\beta=0.4$) shown strictly inside.
Result
For all $\alpha,\beta$ satisfying $0<\alpha<1$, $0<\beta<1$, $0<\alpha+\beta<1$, the point
is a strict convex combination of the vertices $1$, $1+i$, $i$ (all coefficients positive, summing to $1$), and equivalently satisfies $x<1$, $y<1$, $x+y>1$ — so it lies strictly in the interior of the triangle with those vertices. $\blacksquare$