SOLVE THE EQUATION IN THE COMPLEX NUMBER SYSTEM

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 2kπ 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 θ + i sin θ)

Finding the r :

r = √(12 + 02)

r = √1

r = 1

Finding the α :

 α = tan-1(y/x)

α = tan-1(0/1)

α = tan-1(0)

α = 0

z = 1(cos 0° + i sin 0°)

z16=1[cos (2k𝜋+0)+i sin (2k𝜋+0)]16=116cos 2k𝜋6+i sin 2k𝜋6=cos 2k𝜋6+i sin 2k𝜋6Put k=0,1,2,3,4 and 5When k=0=cos 2(0)𝜋6+i sin 2(0)𝜋6=(cos 0+i sin 0)---(1)When k=1=cos 2(1)𝜋6+i sin 2(1)𝜋6=cos 𝜋3+i sin 𝜋3---(2)When k=2=cos 2(2)𝜋6+i sin 2(2)𝜋6=cos 2𝜋3+i sin 2𝜋3---(3)When k=3=cos 2(3)𝜋6+i sin 2(3)𝜋6=(cos 𝜋+i sin 𝜋)---(4)When k=4=cos 2(4)𝜋6+i sin 2(4)𝜋6=cos 4𝜋3+i sin 4𝜋3---(5)=cos 2(5)𝜋6+i sin 2(5)𝜋6=cos 5𝜋3+i sin 5𝜋3

Problem 2 :

x6 + 1 = 0

Solution:

x6 + 1 = 0

x6 = -1

z = -1 + 0i

z = r(cos θ + i sin θ)

Finding the r :

r = √((-1)2 + 02)

r = √1

r = 1

Finding the α :

α = tan-1(y/x)

α = tan-1(0/-1)

α = tan-1(0)

α = 0

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

So, the principal value θ = π - α

θ = π - 0

θ = π 

z = 1(cos π + i sin π)

z16=1[cos (2k𝜋+𝜋)+i sin (2k𝜋+𝜋)]16=116cos 2k𝜋+𝜋6+i sin 2k𝜋+𝜋6=cos 𝜋6(2k+1)+i sin 𝜋6(2k+1)Put k=0,1,2,3,4 and 5When k=0=cos 2(0)𝜋+𝜋6+i sin 2(0)𝜋+𝜋6=cos 𝜋6+i sin 𝜋6---(1)When k=1=cos 2(1)𝜋+𝜋6+i sin 2(1)𝜋+𝜋6=cos 𝜋2+i sin 𝜋2---(2)When k=2=cos 2(2)𝜋+𝜋6+i sin 2(2)𝜋+𝜋6=cos 5𝜋6+i sin 5𝜋6---(3)When k=3=cos 2(3)𝜋+𝜋6+i sin 2(3)𝜋+𝜋6=cos 7𝜋6+i sin 7𝜋6---(4)When k=4=cos 2(4)𝜋+𝜋6+i sin 2(4)𝜋+𝜋6=cos 3𝜋2+i sin 3𝜋2---(5)When k=5=cos 2(5)𝜋+𝜋6+i sin 2(5)𝜋+𝜋6=cos 11𝜋6+i sin 11𝜋6---(6)

Problem 3 :

x4 + 16i = 0

Solution:

x4 + 16i = 0

x4 = -16i

x=4-16i

z = 0 - 16i

z = r(cos θ + i sin θ)

Finding the r :

r = √(02 + (-16)2)

r = √256

r = 16

Finding the α :

α = tan-1(-16/0)

α = tan-1(∞)

α = 3π/2

θ = 3π/2

z=16cos 3𝜋2 + i sin 3𝜋2z1 4=16cos 2k𝜋+3𝜋2+i sin 2k𝜋+3𝜋214Using De Moivre's theorem, bringing the power insidez1 4=1614cos 4k𝜋+3𝜋2+i sin 4k𝜋+3𝜋2=2cos 𝜋2(4k+3)+i sin 𝜋2(4k+3)Put k=0,1,2 and 3When k=0=2cos 𝜋2(4(0)+3)+i sin 𝜋2(4(0)+3)=2cos 3𝜋2+i sin 3𝜋2---(1)When k=1=2cos 𝜋2(4(1)+3)+i sin 𝜋2(4(1)+3)=2cos 7𝜋2+i sin 7𝜋2---(2)When k=2=2cos 𝜋2(4(2)+3)+i sin 𝜋2(4(2)+3)=2cos 11𝜋2+i sin 11𝜋2---(3)When k=3=2cos 𝜋2(4(3)+3)+i sin 𝜋2(4(3)+3)=2cos 15𝜋2+i sin 15𝜋2---(4)
x=532i

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