Question

Express `(2+"i")/((1+"i")(1-2"i")`  in the form of a + ib. Find its modulus and argument

Answer 1

`"z"=(2+"i")/((1+"i")(1-2"i") )`

 

`= (2+"i")/(1-2"i"+"i"-2"i"^2)=(2+"i")/(3-"i")`

 

`"z" =(2+"i")/(3-"i")xx(3+"i")/(3+"i")=((2+"i")(3+"i"))/(9-"i"^2)`

 

`=(6+5"i"+"i"^2)/10    = (5+5"i")/10  =(1+"i")/2`

 

Now,    `"Z"=1/2+"i"/2="a"+"i""b"`

 

⇒     a = `1/2`  , b =`1/2`

 

|Z|  = `sqrt("a"^2+"b"^2)=sqrt(1/4+1/4)=1/sqrt2`

 

`"arg Z" = tan^-1("b"/"a")= tan^-1(1)=(pi/4)`

 

Adding (i) and (ii), we get :

 

`2"I"=int _0^(pi/2) (sin"x"-cos"x"+cos"x"-sin"x")/(1+sin"x"cos"x")"dx"`

 

`2"I" = 0  ⇒ "I" = 0`

 

⇒  `int _0^(pi/2) (sin"x"-cos"x")/(1+sin"x"cos"x")"dx"=0`