Question

18.1) Find the real and imaginary part:
1+7i/(2-i)²​

Answer 1

Answer:

hope it will help.........

Answer 2

Answer:

Real part, a = - 1

Imaginary part, ib = i

Step-by-step explanation:

$\frac{1 + 7i}{(2 - i) {}^{2} } = \frac{1 + 7i}{2 {}^{2} + i {}^{2} - 2(2)(i) } \\ = \frac{1 + 7i}{4 + ( - 1) - 4i} \\ = \frac{1 + 7i}{3 - 4i} \\ = \frac{(1 + 7i)(3 + 4i)}{(3 -4i)(3 + 4i)} \\ = \frac{ (1)(3 + 4i) + 7i(3 + 4i)}{(3) {}^{2} - (4i) {}^{2} } \\ = \frac{3 + 4 i+ 7i(3) + 7i(4i)}{9 - 16i {}^{2} } \\ = \frac{3 + 4i + 21i + 28(i) {}^{2} }{9 - 16( - 1)} \\ = \frac{3 + 25i + 28( - 1)}{9 + 16} \\ = \frac{3 + 25i - 28}{25} \\ = \frac{ 25i - 25}{25} \\ = \frac{25(i - 1)}{25} \\ = (i - 1)$

If a+ib = - 1 + i

Then

• a = - 1
• ib = i