Solve 1/1+i in polar form ?

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Solve 1/1+i in polar form ?

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Answer 1 · 2018-03-28T07:17:54

$\frac{1}{1 + i} = \frac{\sqrt{2}}{2} [cos (- \frac{π}{4}) + i sin (- \frac{π}{4})]$ $1/(1+i)=sqrt2/2[cos(-pi/4)+isin(-pi/4)]$

Explanation:

We are asked to compute

$\frac{1}{1 + i}$ $1/(1+i)$

and express the result in the form $r [cos θ + i sin θ]$ $r[costheta+isintheta]$ .

First convert the numerator and denominator to polar form.

$\frac{1}{1 + i} = \frac{1 [cos (0) + i sin (0)]}{\sqrt{2} [cos (\frac{π}{4}) + i sin (\frac{π}{4})]}$ $1/(1+i)=(1[cos(0)+isin(0)])/(sqrt2[cos(pi/4)+isin(pi/4)])$

Now we can find the result like so:

$\Rightarrow \frac{1}{\sqrt{2}} [cos (0 - \frac{π}{4}) + i sin (0 - \frac{π}{4})]$ $rArr1/sqrt2[cos(0-pi/4)+isin(0-pi/4)]$

$\Rightarrow \frac{\sqrt{2}}{2} [cos (- \frac{π}{4}) + i sin (- \frac{π}{4})]$ $rArrsqrt2/2[cos(-pi/4)+isin(-pi/4)]$

Answer 2 · 2018-03-28T11:36:24

$\frac{1}{\sqrt{2}} (cos (- \frac{1}{4} π) + i sin (- \frac{1}{4} π)$ $1/sqrt(2)(cos(-1/4pi)+isin(-1/4pi)$

Explanation:

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

$\frac{1 \cdot (1 - i)}{(1 + i) (1 - i)} = \frac{1 - i}{1^{2} - i^{2}} = \frac{1 - i}{2} = \frac{1}{2} - \frac{i}{2}$ $(1*(1-i))/((1+i)(1-i))=(1-i)/(1^2-i^2)=(1-i)/2=1/2-i/2$

From here on it is basicly the same...

$r = \sqrt{{(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$ $r=sqrt((1/2)^2+(1/2)^2)=sqrt(2/4)=1/sqrt(2)$
$θ = {tan}^{- 1} (\frac{\frac{1}{2}}{- \frac{1}{2}}) = {tan}^{- 1} (- 1) = - \frac{1}{4} π$ $theta=tan^-1((1/2)/(-1/2))=tan^-1(-1)=-1/4pi$

$\frac{1}{\sqrt{2}} (cos (- \frac{1}{4} π) + i sin (- \frac{1}{4} π)$ $1/sqrt(2)(cos(-1/4pi)+isin(-1/4pi)$

Answer 3 · 2018-03-28T12:50:59

see a solution process below;

Explanation:

$\frac{1}{1} + i$ $1/1+i$

$1$ $1$ in polar form $= 1 (cos 0 + i sin 0)$ $= 1(cos0 + isin0)$

Where;

$1 = r = \sqrt{1 + 0} = 1$ $1=r=sqrt(1+0) =1$

$theta = tan-¹0 =0$

$r=√(1+1) =√2$

$theta = tan-¹(1/1)=45$

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: $cos(-theta)=costheta$

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Solve 1/1+i in polar form ?

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