Question

1.25. Prove the reverse triangle inequality (Proposition 1.7(b)) $|z_1 - z_2| \geq |z_1| - |z_2|$. 1.26. Use the previous exercise to show that $\left| \frac{1}{z^2 - 1} \right| \leq \frac{1}{3}$ for every $z$ on the circle $C[0, 2]$.

          1.25. Prove the reverse triangle inequality (Proposition 1.7(b)) $|z_1 - z_2| \geq |z_1| - |z_2|$.
1.26. Use the previous exercise to show that
$\left| \frac{1}{z^2 - 1} \right| \leq \frac{1}{3}$
for every $z$ on the circle $C[0, 2]$.

1.25. Prove the reverse triangle inequality (Proposition 1.7(b)) |z1 - z2| ≥ |z1| - |z2|.
1.26. Use the previous exercise to show that
| (1)/(z^2 - 1)| ≤(1)/(3)
for every z on the circle C[0, 2].

Added by Emma M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Atul Kumar

Step 1

In other words, we need to prove that |z1 - z2| ≥ |z1| - |z2|. To do this, we can start by expressing z1 and z2 in terms of their real and imaginary parts. Let z1 = x1 + iy1 and z2 = x2 + iy2, where x1, y1, x2, and y2 are real numbers. Now, we can rewrite the Show more…

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1.25. Prove the reverse triangle inequality (Proposition 1.7((b)) |z1 - Z2] |z1] - [Z2] 1.26. Use the previous exercise to show that 1 1 3 for every z on the circle C[0,2]