Question

Reduce (1/1-4i-2/1+i)(3-4i/5+i) to standard form

Answer 1

Answer:

Step-by-step explanation:

(1-4i-2+1)(3-4i/5+i)

3-4i/5+i-12+16/5i^2-4i-6+8/5-2i+3i-4/5i^2+i

-6+3+4/5-16/5-1-4i/5+i-4i-12i+8i/5-2i+3i

-4-11/5+4i/5-20

(-20-11+4i-100i)/5

(-31-96i)/5

-1/5(31-96i)

Answer 2

Answer :

$\longrightarrow\tt{\bigg(\dfrac{1}{1-4i}-\dfrac{2}{1+i}\bigg)\bigg(\dfrac{3-4i}{5+i}\bigg)}$

$\longrightarrow\tt{\bigg[\dfrac{(1+i-2(1-4i)}{(1-4i)(1+i)}\bigg]\bigg(\dfrac{3-4i}{5+i}\bigg)}$

$\longrightarrow\tt{\bigg(\dfrac{1+i-2+8i}{1+i-4i-4{i}^2}\bigg)\bigg(\dfrac{3-4i}{5+i}\bigg)}$

$\longrightarrow\tt{\bigg(\dfrac{-1+9i}{1-3i+4}\bigg)\bigg(\dfrac{3-4i}{5+i}\bigg)}$

$\longrightarrow\tt{\bigg(\dfrac{-1+9i}{5-3i}\bigg)\bigg(\dfrac{3-4i}{5+i}\bigg)}$

$\longrightarrow\tt{\bigg[\dfrac{-3+4i+27i-36{i}^2}{25+5i-15i-3{i}^2}\bigg]}$

$\longrightarrow\tt{\bigg(\dfrac{-3+4i+27i+36}{25-10i+3}\bigg)}$

$\longrightarrow\tt{\dfrac{31i + 33}{28-10i}}$

$\longrightarrow\tt{\dfrac{31i + 33}{28-10i}\times\dfrac{28+10i}{28+10i}}$

$\longrightarrow\tt{\dfrac{924+330i+868i+310{i}^2}{784+280i-280i-100{i}^2}}$

$\longrightarrow\tt{\dfrac{924-310+330i+868i}{784+100}}$

$\longrightarrow\tt{\dfrac{614+1,198i}{884}}$

$\longrightarrow\tt{\dfrac{2(307+599i)}{2\times 442}}$

$\longrightarrow\tt{\dfrac{307+599i}{442}}$

Hence, the standard form is :

$\longrightarrow\bf\underline{\dfrac{307}{442}+\dfrac{599}{442}i}$