If left(frac{1+i}{1-i}right)^m=1,then m canno | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that,$\left(\frac{1+i}{1-i}ight)^m=1$. First,simplify the base: $\frac{1+i}{1-i} = \frac{1+i}{1-i} 	imes \frac{1+i}{1+i} = \frac{(1+i)^2}

Vedclass · Accepted Answer

$1934$

Vedclass · Answer

(A) Given that,$\left(\frac{1+i}{1-i}ight)^m=1$. First,simplify the base: $\frac{1+i}{1-i} = \frac{1+i}{1-i} 	imes \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1+i^2+2i}{1-(-1)} = \frac{1-1+2i}{2} = \frac{2i}{2} = i$. So,the equation becomes $i^m = 1$. We know that $i^n = 1$ if and only if $n$ is a multiple of $4$. Checking the options: $1934 \div 4 = 483.5$ (not a multiple of $4$). $2024 \div 4 = 506$ (multiple of $4$). $2172 \div 4 = 543$ (multiple of $4$). $10^{100} = (2 	imes 5)^{100} = 2^{100} 	imes 5^{100}$,which is divisible by $4$ since $2^{100}$ is divisible by $4$. Thus,$m$ cannot be $1934$.

If $\left(\frac{1+i}{1-i}\right)^m=1$,then $m$ cannot be equal to

Similar Questions

$(1+i)^{2024}+(1-i)^{2024} = $

Express the given complex number in the form $a+ib$: $\left(-2-\frac{1}{3}i\right)^{3}$

If ${i^2} = -1$,then the value of $\sum_{n=1}^{200} {i^n}$ is

In the Argand plane,the quadrant in which $\frac{1+2i}{1-i}$ lies is:

If $\sum\limits_{k = 0}^{100} {{i^k}} = x + iy$,then the values of $x$ and $y$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module