Question

solve for(1/1-2i+3/1+i)(3+4i/2-4i)

Answer 1

[ (1+i) +3(1-2i) / (1-2i) ( 1+i)] * (3+4i/2-4i) {By taking LCM}. = [(1 + i) + (3 - 6i) / 1 + i - 2i - 2i2] [3+4i / 2-4i].

Answer 2

Answer:

$(\frac{1}{1-2i}+\frac{3}{1+i})(\frac{3+4i}{2-4i})=\frac{1}{4}+\frac{9i}{4}$

Step-by-step explanation:

Given : Expression $(\frac{1}{1-2i}+\frac{3}{1+i})(\frac{3+4i}{2-4i})$

To find : Express the given expression in the form of a+ib?

Solution :

$(\frac{1}{1-2i}+\frac{3}{1+i})(\frac{3+4i}{2-4i})$

Solve the first bracket,

$=(\frac{1+i+3-6i}{(1-2i)(1+i)})(\frac{3+4i}{2-4i})$

$=(\frac{4-5i}{1+i-2i-2i^2})(\frac{3+4i}{2-4i})$

$=(\frac{4-5i}{3-i})(\frac{3+4i}{2-4i})$

Multiply both bracket,

$=\frac{(4-5i)(3+4i)}{(3-i)(2-4i)}$

$=\frac{12-15i+16i-20i^2}{6-12i-2i+4i^2}$

$=\frac{32+i}{2-14i}$

Rationalize the denominator,

$=\frac{32+i}{2-14i}\times \frac{2+14i}{2+14i}$

$=\frac{(32+i)(2+14i)}{(2-14i)(2+14i)}$

$=\frac{64+448i+2i+14i^2}{2^2-(14i)^2}$

$=\frac{50+450i}{4+196}$

$=\frac{50+450i}{200}$

$=\frac{50}{200}+\frac{450i}{200}$

$=\frac{1}{4}+\frac{9i}{4}$

Here, $a=\frac{1}{4},b=\frac{9}{4}$

Therefore, $(\frac{1}{1-2i}+\frac{3}{1+i})(\frac{3+4i}{2-4i})=\frac{1}{4}+\frac{9i}{4}$