SOLVE THE EQUATION IN THE COMPLEX NUMBER SYSTEM

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To find the nth root of a complex number, we follow the steps given below.

Step 1 :

Write the given complex number from rectangular form to polar form.

Step 2 :

Add 2kĻ€ to the argument

Step 3 :

Apply De' Moivre's theorem (bring the power to inside)

Step 4 :

Put k = 0, 1, 2, ............ up to n - 1

Solve each equation in the complex number system. Express solutions in polar and rectangular form.

Problem 1 :

x6 - 1 = 0

Solution:

x6 - 1 = 0

x6 = 1

z = 1 + 0i

z = r(cos Īø + i sin Īø)

Finding the r :

r = āˆš(12 + 02)

r = āˆš1

r = 1

Finding the Ī± :

 Ī± = tan-1(y/x)

Ī± = tan-1(0/1)

Ī± = tan-1(0)

Ī± = 0

z = 1(cos 0Ā° + i sin 0Ā°)

z16=1[cos (2kšœ‹+0)+i sin (2kšœ‹+0)]16=116cos 2kšœ‹6+i sin 2kšœ‹6=cos 2kšœ‹6+i sin 2kšœ‹6Put k=0,1,2,3,4 and 5When k=0=cos 2(0)šœ‹6+i sin 2(0)šœ‹6=(cos 0+i sin 0)---(1)When k=1=cos 2(1)šœ‹6+i sin 2(1)šœ‹6=cos šœ‹3+i sin šœ‹3---(2)When k=2=cos 2(2)šœ‹6+i sin 2(2)šœ‹6=cos 2šœ‹3+i sin 2šœ‹3---(3)When k=3=cos 2(3)šœ‹6+i sin 2(3)šœ‹6=(cos šœ‹+i sin šœ‹)---(4)When k=4=cos 2(4)šœ‹6+i sin 2(4)šœ‹6=cos 4šœ‹3+i sin 4šœ‹3---(5)=cos 2(5)šœ‹6+i sin 2(5)šœ‹6=cos 5šœ‹3+i sin 5šœ‹3

Problem 2 :

x6 + 1 = 0

Solution:

x6 + 1 = 0

x6 = -1

z = -1 + 0i

z = r(cos Īø + i sin Īø)

Finding the r :

r = āˆš((-1)2 + 02)

r = āˆš1

r = 1

Finding the Ī± :

Ī± = tan-1(y/x)

Ī± = tan-1(0/-1)

Ī± = tan-1(0)

Ī± = 0

Since, the complex number -1 + 0i is negative and positive, z lies in the second quadrant.

So, the principal value Īø = Ļ€ - Ī±

Īø = Ļ€ - 0

Īø = Ļ€ 

z = 1(cos Ļ€ + i sin Ļ€)

z16=1[cos (2kšœ‹+šœ‹)+i sin (2kšœ‹+šœ‹)]16=116cos 2kšœ‹+šœ‹6+i sin 2kšœ‹+šœ‹6=cos šœ‹6(2k+1)+i sin šœ‹6(2k+1)Put k=0,1,2,3,4 and 5When k=0=cos 2(0)šœ‹+šœ‹6+i sin 2(0)šœ‹+šœ‹6=cos šœ‹6+i sin šœ‹6---(1)When k=1=cos 2(1)šœ‹+šœ‹6+i sin 2(1)šœ‹+šœ‹6=cos šœ‹2+i sin šœ‹2---(2)When k=2=cos 2(2)šœ‹+šœ‹6+i sin 2(2)šœ‹+šœ‹6=cos 5šœ‹6+i sin 5šœ‹6---(3)When k=3=cos 2(3)šœ‹+šœ‹6+i sin 2(3)šœ‹+šœ‹6=cos 7šœ‹6+i sin 7šœ‹6---(4)When k=4=cos 2(4)šœ‹+šœ‹6+i sin 2(4)šœ‹+šœ‹6=cos 3šœ‹2+i sin 3šœ‹2---(5)When k=5=cos 2(5)šœ‹+šœ‹6+i sin 2(5)šœ‹+šœ‹6=cos 11šœ‹6+i sin 11šœ‹6---(6)

Problem 3 :

x4 + 16i = 0

Solution:

x4 + 16i = 0

x4 = -16i

x=4-16i

z = 0 - 16i

z = r(cos Īø + i sin Īø)

Finding the r :

r = āˆš(02 + (-16)2)

r = āˆš256

r = 16

Finding the Ī± :

Ī± = tan-1(-16/0)

Ī± = tan-1(āˆž)

Ī± = 3Ļ€/2

Īø = 3Ļ€/2

z=16cos 3šœ‹2 + i sin 3šœ‹2z1 4=16cos 2kšœ‹+3šœ‹2+i sin 2kšœ‹+3šœ‹214Using De Moivre's theorem, bringing the power insidez1 4=1614cos 4kšœ‹+3šœ‹2+i sin 4kšœ‹+3šœ‹2=2cos šœ‹2(4k+3)+i sin šœ‹2(4k+3)Put k=0,1,2 and 3When k=0=2cos šœ‹2(4(0)+3)+i sin šœ‹2(4(0)+3)=2cos 3šœ‹2+i sin 3šœ‹2---(1)When k=1=2cos šœ‹2(4(1)+3)+i sin šœ‹2(4(1)+3)=2cos 7šœ‹2+i sin 7šœ‹2---(2)When k=2=2cos šœ‹2(4(2)+3)+i sin šœ‹2(4(2)+3)=2cos 11šœ‹2+i sin 11šœ‹2---(3)When k=3=2cos šœ‹2(4(3)+3)+i sin šœ‹2(4(3)+3)=2cos 15šœ‹2+i sin 15šœ‹2---(4)
x=532i

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