Question

1-i/1+I in polar form

Answer 1

So -i is the final form
Now we know that
r=√x^2+y^2
x=0 (since no real part)
y=-1 (Coefficient of I)
hence r=√(-1)^2
r=1
Now theta =tan^-1(y/x)
=tan^-1(-1/0)
=tan^-1(negative infinity)
=-90°
Now polar form
=r(costheta +sintheta)
=1[cos(-90°)+isin(-90°)]
Hope it helps
:)

Answer 2

$z \: = \: \frac{1 \: - \: i}{1 \: + \: i}$
To Rationalize z, we multiple and divide z by (1 – i),
$z \: = \: \frac{(1 \: - \: i)}{(1 \: + \: i)} \: \times \: \frac{(1 \: - \: i)}{(1 \: - \: i)}$
$z \: = \: \frac{{(1 \: - \: i) }^{2} }{(1 \: + \: i)(1 \: - \: i)}$
$z \: = \: \frac{(1 \: - \: 1 \: - \: 2i)}{(1 \: - ( - 1))}$
$z \: = \: \frac{ - 2i}{2}$
$z \: = \: - i$
or,
$z \: = \: 0 \: + \: ( - 1)i$
$a \: = \: 0 \: \: and \: \: b \: = \: - 1$
$r \: = \: |z| \: = \: \sqrt{ {a}^{2} \: + \: {b}^{2} }$
$r \: = \: |z| \: = \: \sqrt{{(0) }^{2} \: + \: {( - 1)}^{2} }$
$r \: = \: 1$
Since, a = 0, b < 0, Graph/Line on Negative Y - axis.
$\tan( \alpha ) \: = \: | \frac{b}{a} |$
$\alpha \: = \: { \tan }^{ - 1} ( | \frac { - 1}{0} | )$

$\alpha = \: {\tan}^{ - 1} ( \frac{1}{0} )$
$\alpha \: = \: {( \frac{ \pi}{2} )}^{ \: c}$
Ø = a – π
Ø = (π/2 – π)
Ø = –π/2
P.F. => r(cosØ + isinØ)
P.F. => 1(cos(-π/2) + isin (-π/2)) = 1(e)^(Ø.i)