Square Roots of Complex Numbers

Square Roots of Real and Imaginary Numbers

Just as we can find powers of a Complex number, we can also find any roots of a Complex number by using their Polar representation.

Taking the square root (or any root) of a Real number is the process of finding a Real number whose square is equal to the original number.

Real numbers have 2 square roots, a positive solution and its negative

Example

\sqrt{64}=(64)^{\frac{1}{2}}=\pm8

An Imaginary number has a  positive and a negative square root

Example

\sqrt{-64}=\pm\sqrt{(8^{2})(i^{2})}=\pm8i

Square Root of a Complex Number z=x+iy

When we want to find the square root of a Complex number, we are looking for a certain other Complex number which, when we square it, gives back the first Complex number as a result.

Complex numbers have 2 square roots, a certain Complex number z=x+iy and its opposite -z=-x-iy

Existence of the Square Root

Example

(2+i)^{2}=(4-1)+(2+2)i=3+4i\implies\sqrt{3+4i}=(3+4i)^{\frac{1}{2}}=(2+i)

(-2-i)^{2}=(4-1)+(2+2)i=3+4i\implies\sqrt{3+4i}=(3+4i)^{\frac{1}{2}}=(-2-i)

This shows the existence of the square roots of a Complex number, but does not actually show the process we would use to solve for them.

Polar Form and the Power Formula

To do that, we need to use the Polar form of the Complex number and the power formula

z^{n}=(r(\cos{\theta}+i\sin{\theta}))^{n}=r^{n}(\cos{n\theta}+i\sin{n\theta})

and not use the restriction that n be an Integer and open up its domain to include the Rational numbers \frac{a}{b}

If we choose n=\frac{1}{2} and use the power formula, we get

z^{\frac{1}{2}}=r^{\frac{1}{2}}(\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}})

Example

z=1+\sqrt{3}i with r=2 and \theta=\frac{\pi}{3}\implies\frac{\theta}{2}=\frac{\pi}{6}

then

z^{\frac{1}{2}}=(1+\sqrt{3}i)^{\frac{1}{2}}=\sqrt{2}(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}})=\sqrt{2}(\frac{\sqrt{3}}{2}+\frac{1}{2})=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}

Square of the Opposite

If we square the opposite of the result -\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}, we should get back 1+\sqrt{3}i

(-\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2})^{2}=(\frac{6}{4}-\frac{2}{4})+(\frac{\sqrt{12}}{4}+\frac{\sqrt{12}}{4})=1+\sqrt{3}i

This shows that the 2 square roots are just opposites of each other z and -z

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nth Root of a Complex Number

We can extend our result for the power n=\frac{1}{2} to include n=\frac{1}{3}, \frac{1}{4} ... by using the formula

z_k=r^{\frac{1}{n}}(\cos{\frac{\theta+2\pik}{n}}+i\sin{\frac{\theta+2\pik}{n}})

where z_k is one root out of the total for k=0, 1, 2 ... n-1

If n=6 then k goes form 0 to 5

The solutions are all located the same distance from the origin and are all separated by the same angle, not necessarily as measured from the positive x-axis, but from the vector representing one root to the vector representing the next root in line.

Geometrically, each nth root of z, z_k, is an equally spaced point that forms the vertices of a regular polygon, i.e. each 5th root of a Complex number forms a vertex of a pentagon.

Example

n=4 and \theta=\pi\implies\frac{\theta}{4}=\frac{\pi}{4} then

z_0=r^{\frac{1}{4}}(\cos{(\frac{\pi}{4}+\frac{2\pi(0)}{4})}+i\sin{(\frac{\pi}{4}+\frac{2\pi(0)}{4})})=r^{\frac{1}{4}}(\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}})

z_1=r^{\frac{1}{4}}(\cos{(\frac{\pi}{4}+\frac{2\pi}{4})}+i\sin{(\frac{\pi}{4}+\frac{2\pi}{4})})=r^{\frac{1}{4}}(\cos{\frac{3\pi}{4}}+i\sin{\frac{3\pi}{4}})

z_2=r^{\frac{1}{4}}(\cos{(\frac{\pi}{4}+\frac{4\pi}{4})}+i\sin{(\frac{\pi}{4}+\frac{4\pi}{4})})=r^{\frac{1}{4}}(\cos{\frac{5\pi}{4}}+i\sin{\frac{5\pi}{4}})

z_3=r^{\frac{1}{4}}(\cos{(\frac{\pi}{4}+\frac{6\pi}{4})}+i\sin{(\frac{\pi}{4}+\frac{6\pi}{4})})=r^{\frac{1}{4}}(\cos{\frac{7\pi}{4}}+i\sin{\frac{7\pi}{4}})

Example

We'll find the cube roots of 0+8i=8i at an angle of \frac{\pi}{2}

(8i)^{\frac{1}{3}}

with r=8 and \theta=\frac{\pi}{2}\implies\frac{\theta}{3}=\frac{\pi}{6}

k=0\implies(8i)^{\frac{1}{3}}=(8)^{\frac{1}{3}}(\cos{\frac{\pi}{6}+0}+i\sin{\frac{\pi}{6}+0})=2(\frac{\sqrt{3}}{2}+\frac{1}{2}i)=\sqrt{3}+i

k=1\implies(8i)^{\frac{1}{3}}=(8)^{\frac{1}{3}}(\cos{\frac{\pi}{6}+\frac{2\pi}{3}}+i\sin{\frac{\pi}{6}+\frac{2\pi}{3})=2(\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}})=2(-\frac{\sqrt{3}}{2}+\frac{1}{2}i)=-\sqrt{3}+i

k=2\implies(8i)^{\frac{1}{3}}=(8)^{\frac{1}{3}}(\cos{\frac{\pi}{3}+\frac{4\pi}{3}}+i\sin{\frac{\pi}{3}+\frac{4\pi}{3}})=2(\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}})=2(0-i)=-2i

Example

The 4th root of z=-4

z^{\frac{1}{4}}=(-4)^{\frac{1}{4}}

with r=4\impliesr^{\frac{1}{4}}=\sqrt{2} and \theta=\pi\implies\frac{\theta}{4}=\frac{\pi}{4}

k=0\implies(-4)^{\frac{1}{4}}=\sqrt{2}(\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}})=\sqrt{2}(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)=1+i

k=1\implies\sqrt{2}(\cos{\frac{\pi}{4}+\frac{2\pi}{4}}+i\sin{\frac{\pi}{4}+\frac{2\pi}{4})=\sqrt{2}(\cos{\frac{3\pi}{4}}+i\sin{\frac{3\pi}{4}})=\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)=-1+i

k=2\implies\sqrt{2}(\cos{\frac{\pi}{4}+\frac{4\pi}{4}}+i\sin{\frac{\pi}{4}+\frac{4\pi}{4}})=\sqrt{2}(\cos{\frac{5\pi}{4}}+i\sin{\frac{5\pi}{4}})=\sqrt{2}(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)=-1-i

k=3\implies\sqrt{2}(\cos{\frac{\pi}{4}+\frac{6\pi}{4}}+i\sin{\frac{\pi}{4}+\frac{6\pi}{4}})=\sqrt{2}(\cos{\frac{7\pi}{4}}+i\sin{\frac{7\pi}{4}})=\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)=1-i

The 5th Roots of -32 in Polar Form

z^5=-32

z^{5}=-32\impliesz=(-32)^{\frac{1}{5}} with r^{\frac{1}{5}}=-2 and \theta=\pi\implies\frac{\theta}{5}=\frac{\pi}{5}

\theta_0=\frac{\pi}{5}

z_0=2e^{\frac{\pii}{5}

\theta_1=\frac{\pi}{5}+\frac{2\pi}{5}=\frac{3\pi}{5}

z_1=2e^{\frac{3\pii}{5}

\theta_2=\frac{\pi}{5}+\frac{4\pi}{5}=\frac{5\pi}{5}=\pi

z_2=2e^{\pii}

\theta_3=\frac{\pi}{5}+\frac{6\pi}{5}=\frac{7\pi}{5}

z_3=2e^{\frac{7\pii}{5}

\theta_4=\frac{\pi}{5}+\frac{8\pi}{5}=\frac{9\pi}{5}

z_4=2e^{\frac{9\pii}{5}

The 8th Roots of Unity

8th Roots of Unity

The solutions to the equation z^{8}=1 are shown in the picture above: 1, -1, i, -i, \pm\frac{\sqrt{2}+\sqrt{2}i}{2}, \pm\frac{\sqrt{2}-\sqrt{2}i}{2}.

They are all spaced at an angle of \frac{\pi}{4} apart as \frac{2\pi}{8}=\frac{\pi}{4}. If there were 6 roots, then the roots would be spread apart by an angle \frac{\pi}{3}.

They all lie on the circumference of the Unit Circle a distance of 1 unit away from the origin.

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Patrick