Find and graph the cube roots of 216i?

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  • 1 month ago

    216i => 216[cos(π/2 + 2nπ) + isin(π/2 + 2nπ)]

    (216i)¹​/³ => (216)¹​/³[cos(π/2 + 2nπ) + isin(π/2 + 2nπ)]¹​/³ 

    i.e. 6[cos(π/6 + 2nπ/3) + isin(π/6 + 2nπ/3)]

    For n = 0 we have:

    6[cos(π/6) + isin(π/6)]

    i.e. 6(√3/2 + i/2)

    or, 3(√3 + i)

    With n = 1 we have:

    6[cos(5π/6) + isin(5π/6)]

    i.e. 6(-√3/2 + i/2)

    or, 3(-√3 + i)

    With n = 2 we have:

    6[cos(3π/2) + isin(3π/2)]

    i.e. 6(0 + -i) => -6i

    Hence, 3 cubic roots are -6i, 3(√3 + i) and 3(-√3 + i) 

    :)>

  • Vaman
    Lv 7
    1 month ago

    Cube root. For this, study 216=3*72= 3*3*24=3*3*3* 8

    The cue root of 216 is 6. For i you write i sin pi/2= exp i(2npi +pi/2). The cue root of 216 is 6 exp i(2npi/3+pi/6)

    Now write one root, n=0.

    r1= root 1= 6 (cos pi/6+i sin pi/6)

    r2 (n=1) =6 (exp i( 2pi/3+pi/6)=6 exp i(150). You can expand and get the value

    r3 (n=2) 6 exp(i(4pi/3+pi/6))= 6 exp (i270)

    These are three roots.

  • 1 month ago

    I'll go first......

    1.  convert your  number into trig. form ,   r CiS @     where  r = your number    [  r will always  be > 0,    if  a + bi  given  then  r = sqrt( a^2 + b^2)   ]    and

    @ = angle from the + x axis counter clockwise   [ Tan @ = b/a  solve for  @  ] 

    NOTE:  CiS  does not mean  Crime Investigative Service , like the tv show...it means  Cosine ( )  + i  Sine (  )

    216i  =  216( cos 90˚ + i sin 90˚ )    or  216 CiS 90˚    some books call this  z

    2.   Use  eqn  for  roots   z ^1/n   =  r^1/n  [  CiS  (   ( @ + 360K ) /  n  )      where  K = 0, 1, 2, ... n - 1      [ in radians  ( @ + 2πK ) / n  ]

    so  K = 0, 1, or  2  here   thus  3  roots    

    216^1/3 = 6         

    @ + 360K     will =   90 + 360*0 = 90     for K = 0

    90 + 360*1 =  450  for  K = 1    

    and  90 +360*2 = 810    for K = 2

    now  divide  by  n,  which is 3  here....  angles  are  90/3 = 30˚   , 450/3 = 150˚ , etc...

    so first answer in trig  form is  6 CiS 30˚ ... I'll let you find the others and then convert them into a + bi  form if needed... I hope you know how to do that.

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