If 1+ι1−ιx=1 then
x=4n,where n is any positive integer.
x=2n,where n is any positive integer.
x=4n+1, where n is any positive integer.
x=2n+1,where n is any positive integer.
To find the value of x:
The given equation is 1+ι1−ιx=1
⇒1+ι1−ιx=1⇒(1+ι)(1+ι)(1−ι)(1+ι)x=1⇒(1+ι)22x=1⇒1+2ι−12x=1⇒2ι2x=1⇒ιx=1⇒ιx=ι4n⇒x=4n
where nis a positive integer.
Hence option A: x=4n,where n is any positive integer.
Explain, (iota) i2 = 1