Question

Complex number ..
Express in the form of a+ib, a,b belongs to R and i =√-1
3+2i/2-5i+3-2i/2+5i

Answer 1

Answer:  The required form is $-\dfrac{8}{29}+i\times 0,$ where $a=-\dfrac{8}{29},~~b=0.$

Step-by-step explanation:  We are given to express the following in the form of a + ib, where a, b are real numbers and i =√-1 :

$C=\dfrac{3+2i}{2-5i}+\dfrac{3-2i}{2+5i}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We will be using the following formulas :

$(i)~(x+y)(x-y)=x^2-y^2,\\\\(ii)~(x+y)(z+w)=x(z+w)+y(z+w).$

From expression (i), we have

$C\\\\=\dfrac{3+2i}{2-5i}+\dfrac{3-2i}{2+5i}\\\\\\=\dfrac{(3+2i)(2+5i)+(3-2i)(2-5i)}{(2-5i)(2+5i)}\\\\\\=\dfrac{6+15i+4i+10i^2+6-15i-4i+10i^2}{2^2-(5i)^2}\\\\\\=\dfrac{12+20i^2}{4-25i^2}\\\\\\=\dfrac{12-20}{4+25}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }i^2=-1]\\\\\\=\dfrac{-8}{29}\\\\\\=-\dfrac{8}{29}+i\times 0.$

Thus, the required form is $-\dfrac{8}{29}+i\times 0,$ where $a=-\dfrac{8}{29},~~b=0.$