- is an imaginary cube root of unity- If 1-2 m -1-4 m then least positive integral value of m isA- 6B- 5C- 4D- 3

We have, (1+ω^2)^m=(1+ω^4)^m (∵ω^3=1) (1+ω^2)^m=(1+ω)^m ( ω)^m=( ω^2)^m ⇒(ω/ω^2 )^m=1⇒ (ω^2)^m=1=(ω)^2m=(ω^3)⇒ m=3/2 Hence least positive integral val ...

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Standard XII

Mathematics

Square Root of a Complex Number

ω is an imagi...

Question

$ω$ is an imaginary cube root of unity. If $(1 + ω^{2})^{m} = (1 + ω^{4})^{m}$ then least positive integral value of m is

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Solution

The correct option is D 3
$W e h a v e, (1 + ω^{2})^{m} = (1 + ω^{4})^{m} (∵ ω^{3} = 1) (1 + ω^{2})^{m} = (1 + ω)^{m} (- ω)^{m} = (- ω^{2})^{m} \Rightarrow {(\frac{ω}{ω^{2}})}^{m} = 1 \Rightarrow (ω^{2})^{m} = 1 = (ω)^{2 m} = (ω^{3}) \Rightarrow m = \frac{3}{2} H e n c e l e a s t p o s i t i v e i n t e g r a l v a l u e o f m i s 3$ .

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ω is an imaginary cube root of unity. If (1+ω2)m=(1+ω4)m then least positive integral value of m is

The correct option is D 3We have,(1+ω2)m=(1+ω4)m (∵ω3=1)(1+ω2)m=(1+ω)m(−ω)m=(−ω2)m⇒(ωω2)m=1⇒(ω2)m=1=(ω)2m=(ω3)⇒m=32Hence least positive integral value of m is 3.

$ω$ is an imaginary cube root of unity. If $(1 + ω^{2})^{m} = (1 + ω^{4})^{m}$ then least positive integral value of m is