If (1+i1-i)^x=1, then

Question

If $\left(\dfrac{1+i}{1-i}\right)^x=1$, then

Accepted Answer

$${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1$$

$$ \Rightarrow {\left( i \right)^x} = 1$$

We know $$i = \sqrt { - 1} $$

$$\therefore$$ ${i^2} =  - 1$

$$ \Rightarrow $$ $${i^3} =  - 1 \times i =  - i$$

$$ \Rightarrow $$ $${i^4} =  - i \times i =  - {i^2} =  - \left( { - 1} \right) = 1$$

So when power of $i$ is 4 or multiple of 4 then it's value is = 1

$$\therefore$$ $${\left( i \right)^x} = 1$$ $$ = {\left( i \right)^{4n}}$$ where n is a positive integer.

Solutions · 1