Question

Solve 1/1+i in polar form ?

Answer 1

`1/(1+i)=sqrt2/2[cos(-pi/4)+isin(-pi/4)]`

Explanation:

We are asked to compute

`1/(1+i)`

and express the result in the form `r[costheta+isintheta]`.

First convert the numerator and denominator to polar form.

`1/(1+i)=(1[cos(0)+isin(0)])/(sqrt2[cos(pi/4)+isin(pi/4)])`

Now we can find the result like so:

`rArr1/sqrt2[cos(0-pi/4)+isin(0-pi/4)]`

`rArrsqrt2/2[cos(-pi/4)+isin(-pi/4)]`

Answer 2

`1/sqrt(2)(cos(-1/4pi)+isin(-1/4pi)`

Explanation:

The method Sharkbasket used is completly right, this is just a second way...
`1/(1+i)->`
`"Expand the fraction by" (1-i)`
`"to get rid of the i in the denominator"`

`(1*(1-i))/((1+i)(1-i))=(1-i)/(1^2-i^2)=(1-i)/2=1/2-i/2`

From here on it is basicly the same...

`r=sqrt((1/2)^2+(1/2)^2)=sqrt(2/4)=1/sqrt(2)`
`theta=tan^-1((1/2)/(-1/2))=tan^-1(-1)=-1/4pi`

`1/sqrt(2)(cos(-1/4pi)+isin(-1/4pi)`

Answer 3

see a solution process below;

Explanation:

`1/1+i`

`1` in polar form `= 1(cos0 + isin0)`

Where;

`1=r=sqrt(1+0) =1`

`theta = tan-¹0 =0`

`r=√(1+1) =√2`

`theta = tan-¹(1/1)=45`

Now;

`1/1+i =[1(cos0 +isin0)]/[sqrt2(cos45 +isin45)]`

By complex analysis we have;

`1/sqrt2[cos(0-45) +isin(0-45)]`

Therefore;

`1/1+i = (sqrt2)/2[cos45 -isin45]`

Note: `cos(-theta)=costheta`