Question

plss. solve this sum step by step​

Answer 1

GIVEN :–

$\\ \:\:\: \left( \dfrac{1}{1 - 2i} + \dfrac{3}{1 + i} \right) \left( \dfrac{3 + 4i}{2 - 4i} \right)$

TO FIND :–

• Simplified value = ?

SOLUTION :–

• Let the complex number –

$\\ \implies \: Z = \: \left( \dfrac{1}{1 - 2i} + \dfrac{3}{1 + i} \right) \left( \dfrac{3 + 4i}{2 - 4i} \right)$

$\\ \implies \: Z = \: \left( \dfrac{1(1 + i) + 3(1 - 2i)}{(1 - 2i)(1 + i)}\right) \left( \dfrac{3 + 4i}{2 - 4i} \right)$

$\\ \implies \: Z = \: \left( \dfrac{1 + i + 3 - 6i}{(1 - 2i)(1 + i)}\right) \left( \dfrac{3 + 4i}{2 - 4i} \right)$

$\\ \implies \: Z = \: \left( \dfrac{4 - 5i}{1 + i - 2i - 2 {i}^{2} }\right) \left( \dfrac{3 + 4i}{2 - 4i} \right)$

$\\ \implies \: Z = \: \left( \dfrac{4 - 5i}{1 + i - 2i - 2( - 1) }\right) \left( \dfrac{3 + 4i}{2 - 4i} \right) \: \: \: \: [ \: \because \: {i}^{2} = - 1]$

$\\ \implies \: Z = \: \left( \dfrac{4 - 5i}{1 + i - 2i + 2}\right) \left( \dfrac{3 + 4i}{2 - 4i} \right)$

$\\ \implies \: Z = \: \left( \dfrac{4 - 5i}{3 - i}\right) \left( \dfrac{3 + 4i}{2 - 4i} \right)$

$\\ \implies \: Z = \: \left( \dfrac{12 + 16i - 15i - 20 {i}^{2}}{ 6 - 12i - 2i + 4 {i}^{2}}\right)$

$\\ \implies \: Z = \: \left( \dfrac{12 + 16i - 15i - 20( - 1)}{ 6 - 12i - 2i + 4( - 1)}\right)$

$\\ \implies \: Z = \: \left( \dfrac{12 + 16i - 15i + 20}{ 6 - 12i - 2i - 4}\right)$

$\\ \implies \: Z = \: \left( \dfrac{32 + i}{ 2- 14i }\right)$

• Now Rationalization –

$\\ \implies \: Z = \: \left( \dfrac{32 + i}{ 2- 14i } \times \dfrac{2 + 14i}{2 + 14i} \right)$

$\\ \implies \: Z = \: \left( \dfrac{64 + 448i + 2i + 14 {i}^{2}}{4 - 196 {i}^{2} } \right)$

$\\ \implies \: Z = \: \left( \dfrac{64 + 448i + 2i + 14( - 1)}{4 - 196( - 1)} \right)$

$\\ \implies \: Z = \: \left( \dfrac{64 + 448i + 2i - 14}{4 + 196} \right)$

$\\ \implies \: Z = \: \left( \dfrac{50 + 450i }{200} \right)$

$\\ \implies \large{ \boxed{Z = \frac{1 + 9i}{4} }} \:$