Imaginary Numbers: Power of i

Complex Numbers :Power of i



We did not define nth root of a certain number a if n is even and a is negative. For instance square root of -9 was not defined . In particular we did not defineSQRT(a) if a is a negative since there is no real number whose square root is negative.Thus we cannot solve an equation such asX^2= -81 using only real numbers.We may extend the real number systemto a larger system called the complex numbers system.To do this we first define the imaginary numberii :



i^2 =- 1or i = SQRT(-1)

Therefore-i =i^3

Sincei^3=( i^2)( i )=( -1) (i ) =-i

i^4 =1

Sincei^4=( i^2) (i^2)=(-1)(-1)=1

i^5 = i

Sincei^5=( i^4)(i ) = (1) (i)=i


Sample Exercises



Number One :Show thati^27=-i

i^27 =(i^24 )(i^2)(i)

Since i^24=( i^4)^6=1^6=1 and i^2=-1

Thereforei^27=(1) (-1)(i)=-i



Number Two:Findi^105

i^105 =(i^104 )(i)

Sincei^104=(i^4)^26 = 1^26=1

Threforei^105= ( 1)(i)= i


Number Three:Find i^307

i^307 =( i^304)( i^2) (i)

Sincei^304=(i^4)^76=1^76= 1 andi^2=-1

Thereforei^307=(1) (-1) (i)=-i



Number Four : Find i^1002

i^1002 = ( i^1000) (i^2)

Since i^1000= ( i^4)^250=1^250=1andi^2= -1

Therefore i^1002=(1) (-1)=-1




SOURCE :


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Comments 2 comments

Lord De Cross profile image

Lord De Cross 5 years ago

Cristina you brought some memories from college, exceptionally explained...hopefully kids will read you and get this usbject trhough. Thanks for your effort in explaining Imaginary numbers.uSEFUL AND VOTED UP!

LORD


cristina327 profile image

cristina327 5 years ago from Manila Author

Hi lord de cross Thank you for taking time to visit this hub. Thank you also for the vote. Your visit is much appreciated. Remain blessed always. Best regards.

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