Question #cbbe3

3 Answers
Alan P.
Dec 29, 2017

There is no minimum value for #m#

Explanation:

#(1+i)/(1-i)=i#

#i^m=1# for all #m=4k, k in ZZ# (including negative values of #k#)

#m in {...-12,-8,-4,0,+4,+8,+12,...}#

For example if #m=-12#
then #((1+i)/(1-i))^(-12)#

#color(white)("XXX")=i^(-12)#

#color(white)("XXX")=1/(i^12)#

#color(white)("XXX")=1/i^4 xx 1/i^4 xx 1/(i^4)#

#color(white)("XXX")=1/(i^2 * i^2) xx 1/(i^2 * i^2) xx 1/(i^2 xx i^2)#

#color(white)("XXX")=1/((-1) * (-1)) xx 1/((-1) * (-1)) xx 1/((-1) * (-1))#

#color(white)("XXX")=1/1 xx 1/1 xx 1/1#

#color(white)("XXX")=1#

Sunayan S
Dec 29, 2017

See the answer below....

Explanation:

#((1+i)/(1-i))^m=1#
#=>{((1+i)(1+i))/((1+i)(1-i))}^m=1#
#=>{(1+i)^2/(1-i^2)}^m=1#
#=>{(1+2i+i^2)/(1-(-1))}^m=1##" "##color(red)([i=sqrt(-1);i^2=-1]#
#=>{(1+2i-1)/(1+1)}^m=1#
#=>((2i)/2)^m=1#
#=>i^m=1#
#=>m=4p#

So, the least value of #m# is #4# when #p# is #1#. [#m# and #p# both are positive integer.]

Hope it helps...
Thank you...

P dilip_k
Dec 29, 2017

#((1+i)/(1-i))^m=1#

#=>((i(1+i))/(i(1-i)))^m=1#

#=>((i(1+i))/(i-i^2))^m=1#

#=>((i(1+i))/(i-(-1)))^m=1#

#=>((i(1+i))/(i+1))^m=1#

#=>i^m=i^(4n)" where " n in ZZ#

Hence #m=4n" where " n in ZZ#

So #m# may be zero or any intger which is multiple of 4. It can not have any minimum value.

But if we impose the condition that m is a positive integer then its minimum value will be 4.

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