Question

Find the two square roots of each complex number by creating and solving polynomial equations.

z = 15 − 8i
z = 8 − 6i
z = −3 + 4i
z = −5 − 12i
z = 21 − 20i
z = 16 − 30i

Answer 1

1) w₁=4 - i w₂= -4 + i

2) w₁= 3 - i w₂= -3 + i

3) w₁= 1 + 2i w₂= - 1 - 2i

4) w₁= 2- 3i w₂= -2 + 3i

5) w₁= 5 - 2i w₂= -5 + 2i

6) w₁= 5 - 3i w₂= -5 + 3i

Explanation:

The root of a complex number is given by:

`\sqrt[n]{z}=\sqrt[n]{r}(Cos((\theta+2k\pi)/(n)) + i Sin((\theta+2k\pi)/(n)))`

where:

r: is the module of the complex number

θ: is the angle of the complex number to the positive axis x

n: index of the root

1) z = 15 − 8i ⇒ r=17 θ= -0.4899 rad

w₁=
`√(17)(Cos((-0.4899)/(2)) + i Sin((-0.4899)/(2)))`=4-i

w₂=
`√(17)(Cos((-0.4899+2\pi)/(2)) + i Sin((-0.4899+2\pi)/(2)))`=-1+i

2) z = 8 − 6i ⇒ r=10 θ= -0.6435 rad

w₁=
`√(10)(Cos(( -0.6435)/(2)) + i Sin(( -0.6435)/(2)))`= 3 - i

w₂=
`√(10)(Cos(( -0.6435+2\pi)/(2)) + i Sin(( -0.6435+2\pi)/(2)))`= -3 + i

3) z = −3 + 4i ⇒ r=5 θ= -0.9316 rad

w₁=
`√(5)(Cos((-0.9316)/(2)) + i Sin((-0.9316)/(2)))`= 1 + 2i

w₂=
`√(5)(Cos((-0.9316+2\pi)/(2)) + i Sin((-0.9316+2\pi)/(2)))`= -1 - 2i

4) z = −5 − 12i ⇒ r=13 θ= 0.4426 rad

w₁=
`√(13)(Cos((0.4426)/(2)) + i Sin((0.4426)/(2)))`= 2- 3i

w₂=
`√(13)(Cos((0.4426+2\pi)/(2)) + i Sin((0.4426+2\pi)/(2)))`= -2 + 3i

5) z = 21 − 20i ⇒ r=29 θ= -0.8098 rad

w₁=
`√(29)(Cos((-0.8098)/(2)) + i Sin((-0.8098)/(2)))`= 5 - 2i

w₂=
`√(29)(Cos((-0.8098+2\pi)/(2)) + i Sin((-0.8098+2\pi)/(2)))`= -5 + 2i

6) z = 16 − 30i ⇒ r=34 θ= -1.0808 rad

w₁=
`√(34)(Cos((-1.0808)/(2)) + i Sin((-1.0808)/(2)))`= 5 - 3i

w₂=
`√(34)(Cos((-1.0808+2\pi)/(2)) + i Sin((-1.0808+2\pi)/(2)))`= -5 + 3i