Question

find the polar form of the complex number 1+I/1-i​

Answer 1

Solution :-

Let,

$\sf z = \dfrac{1 + i}{1 - i}$

Let's write z in the form a + ib.

$: \implies \sf z = \dfrac{(1 + i) \times (1 + i)}{(1 - i) \times (1 + i)}$

$: \implies \sf z = \dfrac{ {(1 + i)}^{2} }{ {1}^{2} - {i}^{2} }$

$: \implies \sf z = \dfrac{ {1}^{2} + {i}^{2} + 2i }{1 - ( -1)}$

$: \implies \sf z = \dfrac{1 - 1 + 2i}{1 + 1}$

$: \implies \sf z = \dfrac{2i}{2}$

$: \implies \underline{\sf z = i}$

$\bf \therefore z = 0 + i$

We know ,

$\sf z = r(cos \theta + isin \theta)$

Equating real and imaginary parts we get,

$\sf rcos \theta=0 \: , \: rsin \theta = 1$

By squaring and adding we get,

$\sf r^{2}(cos^{2} \theta+ sin^{2} \theta = 0^{2}+1^{2}$

$\sf r^{2} \times 1= 1$

$\bf r=1$

 

$\sf rcos \theta =0$

$\implies \sf cos \theta =0$

$\sf rsin \theta =1$

$\implies\sf sin \theta =1$

$\sf \therefore \theta = \dfrac{\pi}{2}$

$\sf \theta$ lies in 1st quadrant.

$\therefore\underline{\bf i=1 \left( cos \dfrac {\pi}{2}+ i \: sin \dfrac {\pi}{2} \right)}$

Answer 2

Solution :-

$\:$

Let ~

$\:$

$\tt\pmb{ z = \frac{1 + i}{1 - i} }$

$\:$

~Lets write z in the form a + ib

$\:$

$\tt \pmb{z = \frac{(1 +i) \times (1 + i) }{(1 - i) \times (1 + i)} }$

$\:$

$\tt \pmb{ z = \frac{(1 + {i})^{2} }{ {1}^{2} - {i}^{2} } }$

$\:$

$\tt \pmb{z = \frac{ {1}^{2} + {i}^{2} + 2i }{1 - ( - 1)} }$

$\:$

$\tt \pmb{z = \frac{1 - 1 + 2i}{1 + 1} }$

$\:$

$\tt \pmb{ z = \frac{2i}{2} }$

$\:$

$\tt \pmb{z = i}$

$\:$

$\tt \pmb{ .°.z = 0 + i}$

$\:$

$\:$

~We know that :-

$\:$

$\tt \pmb{z = r( \cos \theta + i \sin \theta) }$

$\:$

Equating real and imaginary parts we get ~

$\:$

$\tt \pmb{r \cos \theta , \: \sin \theta = 1 }$

$\:$

By Squaring and adding we get~

$\:$

$\tt \pmb{ {r}^{2}( { \cos }^{2} \theta + { \sin }^{2} \theta = {0}^{2} + {1}^{2}) }$

$\:$

$\tt \pmb{ {r}^{2} \times 1 = 1 }$

$\:$

$\tt \pmb{r = 1}$

$\:$

$\:$

$\tt \pmb{r \cos \theta = 0 }$

$\:$

$\implies {\tt { \bold { \cos \theta = 0 }}}$

$\:$

$\tt \pmb{r \sin \theta = 1 }$

$\:$

$\implies \tt{ \sin \theta = 1 }$

$\:$

$\tt \pmb{.°. \theta = \frac{\pi}{2} }$

$\:$

$\tt \pmb{.°. \theta = \frac{\pi}{2} }$

$\:$

$\:$

$\theta$ Lies in 1st quadrant.

$\:$

$\tt \pmb{.°.i = 1( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) }$

$\:$

$\:$