Question

convert the complex number 1+i/1 - I into polar form

Answer 1

1+i/1-i=1+I/1-i ×1+I/1+i=(1+i)^2/1^2-i^2=1+2i+i^2/1+1=1+2i-1/2=2i/2=i z=0+i. Modulus of complex number =r=√z=√a^2+b^2=√0^2+1^2=√1=1. Argument of complex =0+I=r(cos Tita+isin Tita) =0+i=1(costita+isintita). Therefore cos Tita =0 or sin Tita = 1.(+, +) lies on first quadrant. Argument (z) =90°=π/2.Therefore polar form =r(costita+isin Tita) =1(cosπ/2+sinπ/2).

Answer 2

AnsWer :

Cos π/ 2 + i Sin π/2.

SolutioN :

• Complex number given.

$\tt \dagger \: \: \: \: \: z = \dfrac{1 + i}{1 - i}$

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

$\tt : \implies z = \dfrac{1 + i}{1 - i} \times \dfrac{1 + i}{1 + i}$

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

࿊ We Know,

• i² = - 1.

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

$\tt : \implies z = \dfrac{(1 + i)(1 + 1)}{2 }$

$\tt : \implies z = \dfrac{ {1}^{2} + i + i + {i}^{2} }{2 }$

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

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

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

$\tt : \implies z = \dfrac{ \cancel{ 2}i }{ \cancel2 }$

$\tt : \implies z =i.$

࿊ We know,

• Z = a + bi.

$\tt\dagger \: \: \: \: \: \underline{General \: Equation} \: of \: Z = 0 + i .$

$\rule{90}2$

$\tt : \implies r = \sqrt{ {x}^{2} + {y}^{2} }$

࿊ Where as,

• x = 0.
• y = 1.

$\tt : \implies r = \sqrt{ {0}^{2} + {1}^{2} }$

$\tt : \implies r = \sqrt{ 1}$

$\tt : \implies r = \pm1.$

Now,

Let's find Arg with θ.

$\tt \longmapsto \alpha = tan \bigg| \dfrac{y}{x} \bigg|$

$\tt \longmapsto \alpha = tan \bigg| \dfrac{1}{0} \bigg|$

$\tt \longmapsto \alpha = tan \infty$

࿊ We know,

• The value of tan ∞ = tan 90°

$\tt \longmapsto \alpha = tan \ 90 \degree$

$\tt \longmapsto \alpha = tan \dfrac{ \pi}{2}$

$\rule{90}2$

Now,

• The given Complex number lies on first Quadrant

$\tt \longmapsto \theta= \alpha$

$\tt \longmapsto \theta= tan \dfrac{ \pi}{2}$

$\rule{90}2$

☯ We have, Formula.

• Polar form.

$\tt \longmapsto r \,Cos\, \theta +r\,i\,Sin\,\theta.$

$\tt \longmapsto (1) \,Cos\, \dfrac{ \pi}{2} +(1)\,i\,Sin\, \dfrac{ \pi}{2} .$

$\tt \longmapsto Cos\, \dfrac{ \pi}{2} +i\,Sin\, \dfrac{ \pi}{2} .$

✡ Therefore, the polar form of given Complex number is Cos π/ 2 + i Sin π/2.