A+in=√1+i÷1-i,prove that a^2+b^2=1

Question

asked Jun 24, 2022 99.0k views

1 Answer

Dan Sosedoff · Answer 1 · 2022-06-28T06:54:01+0000

Answer with Step-by-step explanation:

We are given that

$a+ib=\sqrt{(1+i)/(1-i)}$

We have to prove that

a^2+b^2=1

$a+ib=\sqrt{((1+i)(1+i))/((1-i)(1+i))}$

Using rationalization property

$a+ib=\sqrt{((1+i)^2)/((1^2-i^2))}$

Using the property

(a+b)(a-b)=a^2-b^2

$a+ib=\sqrt{((1+i)^2)/((1-(-1)))}$

Using

i^2=-1

a+ib=(1+i)/(√(2))

a+ib=(1)/(√(2))+i(1)/(√(2))

By comparing we get

a=(1)/(√(2)), b=(1)/(√(2))

a^2+b^2=((1)/(√(2)))^2+((1)/(√(2)))^2

a^2+b^2=(1)/(2)+(1)/(2)

a^2+b^2=(1+1)/(2)

a^2+b^2=(2)/(2)

a^2+b^2=1

Hence, proved.