a) The 6th root of (−7) are the solution of the equation
z6+7=0 ............(1)
Where z is a complex number.
z6+7=0 ⟹z6=−7=7(−1) ⟹z=761(−1)61Again ,we known that
eπi=cos(π)+isin(π)
=−1
Again,e2πki=1 for all k∈Z
Therefore,
eπie2kπi=−1 ⟹e(2k+1)πi=−1for all k∈Z .
∴z=761e6(2k+1)πiWhere ,k=0,1,2,3,4,5 give the solution of equation (1).
Therefore ,the 6th root of (−7) are
αe6πi,αe63πi,αe65πi,αe67πi,αe69πi,αe611πi
Where α=761 .
The argand diagram is at following photo.
b) Let f(x)=3x−5 ,∀ x∈R
A little algebraic manipulation gives us
∣f(x)−1∣=∣3x−5−1∣=∣3(x−2)∣=3∣x−2∣.
Now , for a given ϵ>0 ,we choose δ=3ϵ . Then if ∣x−2∣<δ,
∣f(x)−1∣=∣3x−5−1∣=∣3x−6∣=3∣x−2∣<3δ=ϵ
Since ϵ>0 is arbitrary, we deduce that limx→2f(x)=1
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