Question

Find the value of
$\left( \dfrac{1}{1-2i}+\dfrac{3}{1+i} \right)\left( \dfrac{3+4i}{2-4i} \right)=$

Answer 1

- Hint: Solve the both brackets separately and then multiply them together.
After multiplying, rationalize the result you got to get a fine answer.
Apply the distributive law:
a.(b + c) = a.b + a.c

Complete step-by-step solution -

We will solve the question step by step,
So first, we will take first bracket,
$\text{= }\left( \dfrac{1}{1-2i}+\dfrac{3}{1+i} \right)$
By taking L.C.M. and solving like algebraic expression, we get:
$=\left( \dfrac{1.\left( 1+i \right)+3.\left( 1-2i \right)}{\left( 1-2i \right).\left( 1+i \right)} \right)$
Now apply the distributive law:
a.(b + c) = a.b + a.c
By applying distributive law, we get:
$=\left( \dfrac{1+i+3-6i}{1.\left( 1+i \right)-2i\left( 1+i \right)} \right)$
By again applying distributive law, we get:
$=\left( \dfrac{1+i+3-6i}{1+i-2i-2{{i}^{2}}} \right)$
By simplifying, we get:
$=\left( \dfrac{4-5i}{1-i-2{{i}^{2}}} \right)$
We know that the variable i is solution of the equation:
${{x}^{2}}=-1$
By substituting this equation, we get:
$=\left( \dfrac{4-5i}{3-i} \right)$
Now we need to multiply this with the next bracket.
$=\left( \dfrac{4-5i}{3-i} \right)\left( \dfrac{3+4i}{2-4i} \right)$
$=\left( \dfrac{\left( 4-5i \right)\left( 3+4i \right)}{\left( 3-i \right)\left( 2-4i \right)} \right)$
Now solve the numerator and denominator separately.
First we take numerator,
$\text{ }\left( 4-5i \right)\left( 3+4i \right)$
By applying distributive law:
a.(b + c) = a.b + a.c
Assume (3 + 4i) as one entity,
Numerator = 4.(3 + 4i) – 5i.(3 + 4i)
By again applying distributive law separately twice, we get:
=(4).(3) + (4).(4i) - (5i).(3) - (5i).(4i)
We know that the variable i is solution of the equation:
${{x}^{2}}=-1$
=12 + 16i – 15i + 20
By simplifying, we get:
Numerator = 32 + i …..(1)
Next we take denominator,
Denominator = (3 - i)(2 - 4i)
By applying distributive law:
a.(b + c) = a.b + a.c
Assume (2 - 4i) as one entity,
Numerator = 3.(2 - 4i) – i.(2 - 4i)
By again applying distributive law separately twice, we get:
=(3).(2) - (3).(4i) - (i).(2) + (i).(4i)
We know that the variable i is solution of the equation:
${{x}^{2}}=-1$
=6 - 12i – 2i - 4
By simplifying, we get:
Denominator = 2 - 14i …..(2)
Now for the required result, divide equation (1) by equation (2).
$\text{= }\dfrac{(1)}{\left( 2 \right)}=\dfrac{\left( 32\text{ }+i \right)}{\left( 2-14i \right)}$
Now rationalize the result.
By multiplying and dividing with (2 – 4i), we get:
$=\dfrac{32+i}{2-14i}.\dfrac{2+14i}{2+14i}$
By applying distributive law:
a.(b + c) = a.b + a.c
By applying above condition, we get:
$=\dfrac{\left( 32\times 2 \right)+\left( 32\times 14i \right)+\left( i\times 2 \right)+\left( i\times 14i \right)}{\left( 2\times 2 \right)+\left( 2\times 14i \right)+\left( -4i\times 2 \right)+\left( -4i\times 14i \right)}$
By simplifying, we get:
$=\dfrac{64+448i+2i-14}{4+28i-28i+196}$
By solving, we get:
$=\dfrac{50+450i}{200}$
By separating the terms, we get:
$=\dfrac{50}{200}+\dfrac{450i}{200}$
By solving, we get:
$=\dfrac{1}{4}+\dfrac{9i}{4}$
$\therefore \left( \dfrac{1}{1-2i}+\dfrac{3}{1+i} \right)\left( \dfrac{3+4i}{2-4i} \right)=\dfrac{1}{4}+\dfrac{9i}{4}$

Note: While solving please do take care of signs as that may change the answer a lot.
Alternative method -
Instead of using distributive law you can use the following identity, while rationalizing.
$\left( a+b \right).\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
So here 2 – 14i is multiplied with 2 + 14i so a = 2 and b = 14i. So the rationalizing can be done in a single step in this method. You get 4 – (196 (i.i)) so by using relation i.i = -1, you get the above value to be 4 + 196 = 200. You got the same result while using distributive law also, but by using this identity we get the result in a single step.