Question

Решите !
1)для чисел z1 2+5i z2=1-7i найдите действительные числа a и b, равенство z1/z2=az1+bz2.
2)запишите z в форме z 13+12i/6i-8 2i+1 2/i+2.​

Answer 1

$\dfrac{z_1}{z_2}=\dfrac{2+5i}{1-7i}=\dfrac{(2+5i)(1+7i)}{(1-7i)(1+7i)}=\dfrac{2+19i-35}{1+49}=\dfrac{19i-33}{50}=\dfrac{19}{50}i-\dfrac{33}{50}\\ \\ az_1+bz_2=a(2+5i)+b(1-7i)=2a+b+i(5a-7b)\\ \\ 2a+b+(5a-7b)i=\dfrac{19}{50}i-\dfrac{33}{50}$

Приравниваем коэффициенты при действительных и мнимых частей.

$\displaystyle \left \{ {{2a+b=-\dfrac{33}{50}} \atop {5a-7b=\dfrac{19}{50}}} \right.~~~\Rightarrow~~~\left \{ {{b=-2a-\dfrac{33}{50}} \atop {5a+7\cdot\left(2a+\dfrac{33}{50}\right)=\dfrac{19}{50}}} \right.\\ \\ \\ 5a+14a+\dfrac{231}{50}=\dfrac{19}{50}\\ \\ 19a=-\dfrac{212}{50}\\ \\ a=-\dfrac{212}{950};\\ \\ \\ b=-2\cdot \left(-\dfrac{212}{950}\right)-\dfrac{33}{50}=\dfrac{424}{950}-\dfrac{627}{950}=-\dfrac{203}{950}$

a = -212/950; b = -203/950

$z=\dfrac{13+12i}{6i-8}-\dfrac{(2i+1)^2}{i+2}=\dfrac{(13+12i)(6i+8)}{(6i-8)(6i+8)}-\dfrac{(-4+4i+1)(i-2)}{(i+2)(i-2)}=\\ \\ \\ =\dfrac{78i+104-72+96i}{-36-64}-\dfrac{(4i-3)(i-2)}{-1-4}=\dfrac{174i+32}{-100}-\dfrac{-4-11i+6}{-5}=\\ \\ \\ =-\dfrac{87i+16}{50}+\dfrac{2-11i}{5}=\dfrac{-87i-16+20-110i}{50}=\dfrac{4-197i}{50}=\dfrac{2}{25}-\dfrac{197}{50}i$