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15. Roots of complex numbers example ( Enter your problem )
  1. Example-1
Other related methods
  1. Adding complex numbers
  2. Subtracting complex numbers
  3. Multiplying complex numbers
  4. Dividing complex numbers
  5. Real part of complex number
  6. Imaginary part of complex number
  7. Conjugate of complex number
  8. Magnitude (Modulus) of complex numbers
  9. Multiplicative inverse of complex numbers
  10. Reciprocal of complex number
  11. Argument of complex number
  12. Polar form of complex numbers
  13. Square root of complex number
  14. Powers of complex numbers
  15. Roots of complex numbers

14. Powers of complex numbers
(Previous method)

1. Example-1





1. `A=-1-i,B=-2+3i,C=1-3i`
Find roots(4,A)


Solution:
Here `A=-1-i,B=-2+3i,C=1-3i`


For a complex number `z=a+bi`, the polar form is `z=r*(cos(theta)+i*sin(theta))`

then all `n^(th)` roots of given complex number can be obtained by the De Moivre's Formula i.e.

`w_k=[r*(cos(theta)+i*sin(theta))]^(1/n)=root (n)(r)*[cos((theta+2kpi)/n)+i*sin((theta+2kpi)/n)]`, where `k=0,1,2..,n-1`


Here `n=4`

Step-1: Convert to polar form: `z=r*(cos(theta)+i*sin(theta))`

Here, `a=-1` and `b=-1`

`:. r=sqrt((-1)^2+(-1)^2)=sqrt(1+1)=sqrt(2)=1.4142`

`theta=atan(b/a)+180` (Since `a<0`)

`:. theta=atan((-1)/(-1))+180`

`:. theta=atan(1)+180`

`:. theta=45+180`

`:. theta=225 ^circ` or `theta=(5pi)/(4)` rad = 3.927 rad

`:. theta=3.927`

The polar form is `r*(cos(theta)+i*sin(theta))`

`=1.4142*(cos(3.927)+i*sin(3.927))`

Step-2: All `n^(th)` roots of given complex number `r*(cos(theta)+i*sin(theta))` is

`w_k=root (n)(r)*[cos((theta+2kpi)/n)+i*sin((theta+2kpi)/n)]`, `k=0,..,n-1`


Hence the 4 roots are
for `k=0`

`w_0=1.0905*[cos((3.927+2*0*pi)/4)+i*sin((3.927+2*0*pi)/4)]`

`w_0=1.0905*[cos(0.9817)+i*sin(0.9817)]`

`w_0=0.6059+0.9067i`


for `k=1`

`w_1=1.0905*[cos((3.927+2*1*pi)/4)+i*sin((3.927+2*1*pi)/4)]`

`w_1=1.0905*[cos(2.5525)+i*sin(2.5525)]`

`w_1=-0.9067+0.6059i`


for `k=2`

`w_2=1.0905*[cos((3.927+2*2*pi)/4)+i*sin((3.927+2*2*pi)/4)]`

`w_2=1.0905*[cos(4.1233)+i*sin(4.1233)]`

`w_2=-0.6059-0.9067i`


for `k=3`

`w_3=1.0905*[cos((3.927+2*3*pi)/4)+i*sin((3.927+2*3*pi)/4)]`

`w_3=1.0905*[cos(5.6941)+i*sin(5.6941)]`

`w_3=0.9067-0.6059i`
2. `A=5+6i,B=-2+3i,C=1-3i`
Find roots(4,A)


Solution:
Here `A=5+6i,B=-2+3i,C=1-3i`


For a complex number `z=a+bi`, the polar form is `z=r*(cos(theta)+i*sin(theta))`

then all `n^(th)` roots of given complex number can be obtained by the De Moivre's Formula i.e.

`w_k=[r*(cos(theta)+i*sin(theta))]^(1/n)=root (n)(r)*[cos((theta+2kpi)/n)+i*sin((theta+2kpi)/n)]`, where `k=0,1,2..,n-1`


Here `n=4`

Step-1: Convert to polar form: `z=r*(cos(theta)+i*sin(theta))`

Here, `a=5` and `b=6`

`:. r=sqrt(5^2+6^2)=sqrt(25+36)=sqrt(61)=7.8102`

`theta=atan(b/a)` (Since `a>0`)

`:. theta=atan((6)/(5))`

`:. theta=atan(1.2)`

`:. theta=50.1944 ^circ` or `theta=0.8761` rad

`:. theta=0.8761`

The polar form is `r*(cos(theta)+i*sin(theta))`

`=7.8102*(cos(0.8761)+i*sin(0.8761))`

Step-2: All `n^(th)` roots of given complex number `r*(cos(theta)+i*sin(theta))` is

`w_k=root (n)(r)*[cos((theta+2kpi)/n)+i*sin((theta+2kpi)/n)]`, `k=0,..,n-1`


Hence the 4 roots are
for `k=0`

`w_0=1.6717*[cos((0.8761+2*0*pi)/4)+i*sin((0.8761+2*0*pi)/4)]`

`w_0=1.6717*[cos(0.219)+i*sin(0.219)]`

`w_0=1.6318+0.3632i`


for `k=1`

`w_1=1.6717*[cos((0.8761+2*1*pi)/4)+i*sin((0.8761+2*1*pi)/4)]`

`w_1=1.6717*[cos(1.7898)+i*sin(1.7898)]`

`w_1=-0.3632+1.6318i`


for `k=2`

`w_2=1.6717*[cos((0.8761+2*2*pi)/4)+i*sin((0.8761+2*2*pi)/4)]`

`w_2=1.6717*[cos(3.3606)+i*sin(3.3606)]`

`w_2=-1.6318-0.3632i`


for `k=3`

`w_3=1.6717*[cos((0.8761+2*3*pi)/4)+i*sin((0.8761+2*3*pi)/4)]`

`w_3=1.6717*[cos(4.9314)+i*sin(4.9314)]`

`w_3=0.3632-1.6318i`
3. `A=5+6i,B=-2+3i,C=1-3i`
Find roots(3,B)


Solution:
Here `A=5+6i,B=-2+3i,C=1-3i`


For a complex number `z=a+bi`, the polar form is `z=r*(cos(theta)+i*sin(theta))`

then all `n^(th)` roots of given complex number can be obtained by the De Moivre's Formula i.e.

`w_k=[r*(cos(theta)+i*sin(theta))]^(1/n)=root (n)(r)*[cos((theta+2kpi)/n)+i*sin((theta+2kpi)/n)]`, where `k=0,1,2..,n-1`


Here `n=3`

Step-1: Convert to polar form: `z=r*(cos(theta)+i*sin(theta))`

Here, `a=-2` and `b=3`

`:. r=sqrt((-2)^2+3^2)=sqrt(4+9)=sqrt(13)=3.6056`

`theta=atan(b/a)+180` (Since `a<0`)

`:. theta=atan((3)/(-2))+180`

`:. theta=atan(-1.5)+180`

`:. theta=-56.3099+180`

`:. theta=123.6901 ^circ` or `theta=2.1588` rad

`:. theta=2.1588`

The polar form is `r*(cos(theta)+i*sin(theta))`

`=3.6056*(cos(2.1588)+i*sin(2.1588))`

Step-2: All `n^(th)` roots of given complex number `r*(cos(theta)+i*sin(theta))` is

`w_k=root (n)(r)*[cos((theta+2kpi)/n)+i*sin((theta+2kpi)/n)]`, `k=0,..,n-1`


Hence the 3 roots are
for `k=0`

`w_0=1.5334*[cos((2.1588+2*0*pi)/3)+i*sin((2.1588+2*0*pi)/3)]`

`w_0=1.5334*[cos(0.7196)+i*sin(0.7196)]`

`w_0=1.1532+1.0106i`


for `k=1`

`w_1=1.5334*[cos((2.1588+2*1*pi)/3)+i*sin((2.1588+2*1*pi)/3)]`

`w_1=1.5334*[cos(2.814)+i*sin(2.814)]`

`w_1=-1.4519+0.4934i`


for `k=2`

`w_2=1.5334*[cos((2.1588+2*2*pi)/3)+i*sin((2.1588+2*2*pi)/3)]`

`w_2=1.5334*[cos(4.9084)+i*sin(4.9084)]`

`w_2=0.2986-1.504i`


This material is intended as a summary. Use your textbook for detail explanation.
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14. Powers of complex numbers
(Previous method)





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