Question

8+5i/4-2i represent in x+yi form

Answer 1

Given,

$\frac{8+5i}{4-2i}$

Multiply and divide $(4+2i)$ with both denominator and numerator

$\frac{8+5i}{4-2i}\times\frac{4+2i}{4+2i}$

The denominator is in the form of $(a-b)(a+b)$,  we can write as $a^{2}-b^{2}$

$\frac{(8+5i)(4+2i)}{4^{2}-2i^{2} }$

$\frac{32+16i+20i+10i^{2} }{16-4i^{2} }$

$\frac{32+36i-10}{16+4}$       ($i^{2}=-1$)

$\frac{36i+22}{20}$

$\frac{22}{20}+\frac{32}{20}i$

$\frac{11}{10}+\frac{9}{5}i$

Therefore, $\frac{8+5i}{4-2i}$ in $x+yi$ form is $\frac{11}{10}+\frac{9}{5}i$.

Answer 2

Given,

$\frac{8+5i}{4-2i}$

To solve the given equation,

We have to multiply and divide the given with $4+2i$

We get,

$=\frac{8+5i}{4-2i} \times \frac{4+2i}{4+2i}$

$=\frac{(8+5i)(4+2i)}{(4-2i)(4+2i)}$

Denominator is in the form of $(a+b)(a-b)$ form we can write it as $(a+b)^{2}$

$=\frac{32+16i+20i+10i^{2} }{16-4i^{2} }$

$=\frac{32+36i-10}{16+4}$

$=\frac{36i+22}{20}$

$=\frac{22}{10}+i\frac{36}{20}$

$=\frac{11}{10} +i \frac{9}{5}$

Therefore, the final answer is $=\frac{11}{10} +i \frac{9}{5}$