Question

solve it step by step plzz 16th and 17 th one​

Answer 1

$\underline{\sf Explanation:}$

Q1)

$\sf z = \frac{5}{(1-i) (2-i) (3-i)}$

$\longrightarrow\sf \frac{5}{(1-i) (2-i) (3-i)}$

$\longrightarrow \sf \frac{5}{(2 - i - 2i + i^2 )(3 - i)}$

$\longrightarrow \sf \frac{5}{(2 - 3i - 1)(3 - i)}$

$\longrightarrow\sf \frac{5}{(1- 3i) (3 - i) }$

$\longrightarrow \sf \frac{5}{3-i-9i+3i^2 }$

$\longrightarrow\sf \frac{5}{3 - 10i - 3}$

$\longrightarrow\sf \frac{5}{-10i}$

$\longrightarrow\sf\frac{1}{-2i}$

multiply and divide by i

$\longrightarrow\sf\frac{1 \times i }{-2i \times i }$

$\longrightarrow\sf\frac{1i}{-2i^2}$

$\longrightarrow\sf\frac{1i}{2}$

therefore ($\sf\frac{1}{2}i$ )is a purely imaginary number.

____________________________

Q2)

$\sf \frac{x}{(1+2i)}+\frac{y}{(3+2i)} = \frac{(5+6i)}{(-1+8i)}$

$\longrightarrow\sf\frac{x(3+2i)+y(1+2i)}{(1+2i)(3+2i)}=\frac{(5+6i)}{(-1+8i)}$

$\longrightarrow\sf\frac{3x+2xi+y+2yi}{3+2i+6i+4i^2} = \frac{5+6i}{-1+8i}$

$\longrightarrow\sf \frac{(3x + y) + i(2x + 2y) }{3+8i-4} = (\frac{5+6i}{-1+8i})$

$\longrightarrow\sf\frac{(3x + y) + i (2x + 2y) }{-1 + 8i} = (\frac{5+6i}{-1+8i})$

$\longrightarrow\sf (3x + y) + i(2x + 2y) = 5 + 6i$

Now, compare the real and imaginary values,

$\sf 3x + y = 5$

and $\sf 2x + 2y = 6$ ---(1)

multiply $\sf 3x + y = 5$ by 2

$\longrightarrow\sf 6x + 2y = 10$ -----(2)

subtract equation (1) from (2)

6x + 2y = 10

2x + 2y = 6

(-)-----(-)-----(-)

______________

4x = 4

$\longrightarrow\sf x = 1$

put the value of x in equation (1)

$\longrightarrow\sf 2x + 2y = 6$

$\longrightarrow\sf 2(1)+2y = 6$

$\longrightarrow\sf 2 + 2y = 6$

$\longrightarrow\sf y = (\frac{6 - 2}{2})$

Hence y = 2

therefore the values of x and y are 1 and 2