Question

Sa se determine modulul numarului complex z=5+3i supra 3+5i .
​

Answer 1

 

$\displaystyle\bf\\z=\frac{5+3i}{3+5i}\\\\Rationalizam~fractia.\\\\z=\frac{(5+3i)(3-5i)}{(3+5i)(3-5i)}\\\\Desfacem~parantezele.\\\\z=\frac{15+9i-25i-15i^2}{9-25i^2}\\\\i^2=-1\\\\z=\frac{15+9i-25i-15\times(-1)}{9-25\times(-1)}\\\\z=\frac{15+9i-25i+15}{9+25}\\\\z=\frac{30-16i^{(2}}{34}\\\\z=\frac{15-8i}{17}\\\\z=\frac{15}{17}-\frac{8}{17}i$

.

$\displaystyle\bf\\Calculam~modulul.\\\\|z|=\sqrt{\left(\frac{15}{17}\right)^2+\left(-\frac{8}{17}\right)^2}\\\\Minusul~dispare~deoarece~exponentul~este~par.\\\\|z|=\sqrt{\frac{15^2}{17^2}+\frac{8^2}{17^2}}\\\\|z|=\sqrt{\frac{15^2+8^2}{17^2}}\\\\|z|=\sqrt{\frac{225+64}{289}}=\sqrt{\frac{289}{289}}=\frac{\sqrt{289}}{\sqrt{289}} =\frac{17}{17}=1\\\\\boxed{\bf |z|=1}$