Find the smallest positive integer n- for which 1-i-1-in-1-

( 1+i/1 i )^n=1 ⇒ ( 1+i/1 i,1+i/1+i )^n=1 ⇒ ( 1+2i+i^2/1 i^2 ) ⇒ ( 1+2i 1/1 ( 1) )^n =1 ⇒ ( 2i/2 )^n=1 ⇒ r^n=1 Now, we know that i^=i, i^2= 1, i^3 = i and i^4 = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Iota

Find the smal...

Question

Find the smallest positive integer n, for which ${(\frac{1 + i}{1 - i})}^{n} = 1.$

Open in App

Solution

${(\frac{1 + i}{1 - i})}^{n} = 1 \Rightarrow {(\frac{1 + i}{1 - i}, \frac{1 + i}{1 + i})}^{n} = 1 \Rightarrow (\frac{1 + 2 i + i^{2}}{1 - i^{2}}) \Rightarrow {(\frac{1 + 2 i - 1}{1 - (- 1)})}^{n} = 1 \Rightarrow {(\frac{2 i}{2})}^{n} = 1 \Rightarrow r^{n} = 1$
Now, we know that $i^{=} i, i^{2} = - 1, i^{3} = - i$ and $i^{4} = 1.$
Hence, the smallest positive integral value of n =4.

Suggest Corrections

Similar questions

Q. The smallest positive integer

n

for which

{\frac{(1 + i)}{(1 - i)}}^{n} = 1

Q. Find the smallest positive integer value of

n

for which

\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}

is a real number.

Q. The smallest positive integer

n

for which

{(\frac{1 + i}{1 - i})}^{n} = 1

where

i = \sqrt{- 1}

Q. The smallest positive integers

n

for which

(\frac{1 + i}{1 - i})^{n} = - 1

is-

Q. The smallest positive integer n for which

(1 + i)^{2 n} = (1 - i)^{2 n}

Join BYJU'S Learning Program

Find the smallest positive integer n, for which (1+i1−i)n=1.

(1+i1−i)n=1⇒(1+i1−i,1+i1+i)n=1⇒(1+2i+i21−i2)⇒(1+2i−11−(−1))n=1⇒(2i2)n=1⇒rn=1 Now, we know that i=i,i2=−1,i3=−i and i4=1. Hence, the smallest positive integral value of n =4.

Find the smallest positive integer n, for which ${(\frac{1 + i}{1 - i})}^{n} = 1.$