Trigonometry - De Moivre's Theorem and Finding Roots of Complex Numbers

Find all fifth roots of .

$\k=0: \sqrt{5}\left (\cos10.63^\circ+i\sin10.63^\circ \right )=R_1\ k=1: \sqrt{5}\left (\cos82.63^\circ+i\sin82.63^\circ \right )=R_2\ k=2: \sqrt{5}\left (\cos154.63^\circ+i\sin154.63^\circ \right )=R_3\ k=3: \sqrt{5}\left (\cos226.63^\circ+i\sin226.63^\circ \right )=R_4\ k=4: \sqrt{5}\left (\cos298.63^\circ+i\sin298.63^\circ \right )=R_5$

$\k=0: \sqrt{5}\left (\cos53.13^\circ+i\sin53.13^\circ \right )=R_1\ k=1: \sqrt{5}\left (\cos125.13^\circ+i\sin125.13^\circ \right )=R_2\ k=2: \sqrt{5}\left (\cos197.13^\circ+i\sin197.13^\circ \right )=R_3\ k=3: \sqrt{5}\left (\cos269.13^\circ+i\sin269.13^\circ \right )=R_4\ k=4: \sqrt{5}\left (\cos341.13^\circ+i\sin341.13^\circ \right )=R_5$

$\k=0: \sqrt[5]{5}\left (\cos10.63^\circ+i\sin10.63^\circ \right )=R_1\ k=1: \sqrt[5]{5}\left (\cos82.63^\circ+i\sin82.63^\circ \right )=R_2\ k=2: \sqrt[5]{5}\left (\cos154.63^\circ+i\sin154.63^\circ \right )=R_3\ k=3: \sqrt[5]{5}\left (\cos226.63^\circ+i\sin226.63^\circ \right )=R_4\ k=4: \sqrt[5]{5}\left (\cos298.63^\circ+i\sin298.63^\circ \right )=R_5$

$\k=0: \sqrt[5]{5}\left (\cos53.13^\circ+i\sin53.13^\circ \right )=R_1\ k=1: \sqrt[5]{5}\left (\cos125.13^\circ+i\sin125.13^\circ \right )=R_2\ k=2: \sqrt[5]{5}\left (\cos197.13^\circ+i\sin197.13^\circ \right )=R_3\ k=3: \sqrt[5]{5}\left (\cos269.13^\circ+i\sin269.13^\circ \right )=R_4\ k=4: \sqrt[5]{5}\left (\cos341.13^\circ+i\sin341.13^\circ \right )=R_5$

Explanation

Begin by converting the complex number to polar form:

$3+4i=5(\cos53.13^\circ+i\sin53.13^\circ)$

Next, put this in its generalized form, using k which is any integer, including zero:

$5[\cos(53.13^\circ+k360^\circ)+i\sin(53.13^\circ+k360^\circ)]$

Using De Moivre's theorem, a fifth root of is given by:

$\left [5[\cos(53.13^\circ+k360^\circ)+i\sin(53.13^\circ+k360^\circ)] \right ]^\frac{1}{5}$

$=5^\frac{1}{5}\left (\cos\frac{53.13^\circ+k360^\circ}{5}+i\sin\frac{53.13^\circ+k360^\circ}{5} \right )$

$=\sqrt[5]{5}\left [\cos(10.63^\circ+k72^\circ)+i\sin(10.63^\circ+k72^\circ) \right ]$

Assigning the values will allow us to find the following roots. In general, use the values .

$k=0: \sqrt[5]{5}\left (\cos10.63^\circ+i\sin10.63^\circ \right )=R_1$

$k=1: \sqrt[5]{5}\left (\cos82.63^\circ+i\sin82.63^\circ \right )=R_2$

$k=2: \sqrt[5]{5}\left (\cos154.63^\circ+i\sin154.63^\circ \right )=R_3$

$k=3: \sqrt[5]{5}\left (\cos226.63^\circ+i\sin226.63^\circ \right )=R_4$

$k=4: \sqrt[5]{5}\left (\cos298.63^\circ+i\sin298.63^\circ \right )=R_5$

These are the fifth roots of .

Simplify using De Moivre's Theorem:

$\small (\cos x+i\sin x)^3$

$\small \small \cos 3x+i\sin3x$

$\small \small 3\cos x+3i\sin x$

$\small \small \sin 3x+i\cos 3x$

$\small \small \cos 3x+i\sin x$

Explanation

We can use DeMoivre's formula which states:

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

Now plugging in our values of $\small n=3$ and we get the desired result.

$\small (\cos x+i\sin x)^n=(\cos nx+i\sin nx)$

$\cos 3x +i \sin 3x$

Find all fourth roots of $-8-8i\sqrt{3}$ .

$\R_1=-1+i\sqrt{3}\ R_2=\sqrt{3}+i\ R_3=1-i\sqrt{3}\ R_4=-\sqrt{3}-i$

$\R_1=1-i\sqrt{3}\ R_2=-\sqrt{3}-i\ R_3=-1+i\sqrt{3}\ R_4=\sqrt{3}+i$

$\R_1=-1-i\sqrt{3}\ R_2=\sqrt{3}-i\ R_3=1+i\sqrt{3}\ R_4=-\sqrt{3}+i$

$\R_1=1+i\sqrt{3}\ R_2=-\sqrt{3}+i\ R_3=-1-i\sqrt{3}\ R_4=\sqrt{3}-i$

Explanation

Begin by converting the complex number to polar form:

$-8-8i\sqrt{3}=16(\cos240^\circ+i\sin240^\circ)$

Next, put this in its generalized form, using k which is any integer, including zero:

$16[\cos(240^\circ+k360^\circ)+i\sin(240^\circ+k360^\circ)]$

Using De Moivre's theorem, a fifth root of $-8-8i\sqrt{3}$ is given by:

$\left [16\cos(240^\circ+k360^\circ)+i\sin(240^\circ+k360^\circ) \right ]^\frac{1}{4}$

$=16^\frac{1}{4}\left [\cos\left (\frac{240^\circ}{4}+\frac{k360^\circ}{4} \right )+i\sin\left (\frac{240^\circ}{4}+\frac{k360^\circ}{4} \right ) \right ]$

$=2\left [\cos\left (60^\circ+k90^\circ \right )+i\sin\left (60^\circ+k90^\circ \right ) \right ]$

Assigning the values will allow us to find the following roots. In general, use the values .

$\k=0: 2\left (\cos60^\circ+i\sin60^\circ \right )=2\left ( \frac{1}{2}+i\frac{1}{2}\sqrt{3} \right )=1+i\sqrt{3}=R_1\ k=1: 2\left (\cos150^\circ+i\sin150^\circ \right )=2\left ( -\frac{1}{2}\sqrt{3}+\frac{1}{2}i \right )=-\sqrt{3}+i=R_2\ k=2: 2\left (\cos240^\circ+i\sin240^\circ \right )=2\left ( -\frac{1}{2}-i\frac{1}{2}\sqrt{3} \right )=-1-i\sqrt{3}=R_3\ k=3: 2\left (\cos330^\circ+i\sin330^\circ \right )=2\left ( \frac{1}{2}\sqrt{3}-\frac{1}{2}i \right )=\sqrt{3}-i=R_4$

These are the fifth roots of $-8-8i\sqrt{3}$ .

Find all cube roots of 1.

$\R_1=1\R_2=-\frac{1}{2}+i\frac{1}{2}\sqrt{3}\R_3=-\frac{1}{2}-i\frac{1}{2}\sqrt{3}$

$\R_1=1\R_2=-\frac{1}{2}+i\frac{1}{2}\R_3=-\frac{1}{2}-i\frac{1}{2}$

$\R_1=1\R_2=-\frac{1}{2}+i\sqrt{3}\R_3=-\frac{1}{2}-i\sqrt{3}$

$\R_1=1\R_2=-\frac{\sqrt{3}}{2}+i\frac{1}{2}\sqrt{3}\R_3=-\frac{\sqrt{3}}{2}-i\frac{1}{2}\sqrt{3}$

Explanation

Begin by converting the complex number to polar form:

$1=\cos0^\circ+i\sin0^\circ$

Next, put this in its generalized form, using k which is any integer, including zero:

$1=\cos(0^\circ+k360^\circ)+i\sin(0^\circ+k360^\circ)$

Using De Moivre's theorem, a fifth root of 1 is given by:

$[\cos(k360^\circ)+i\sin(k360^\circ)] ^\frac{1}{3}$

$=1^\frac{1}{3}\cdot \left [\cos\left (\frac{k360^\circ}{3} \right )+i\sin\left (\frac{k360^\circ}{3} \right ) \right ]$

$=\cos k 120^\circ+i\sin k120^\circ$

Assigning the values will allow us to find the following roots. In general, use the values .

$k=0: \cos0^\circ+i\sin0^\circ=1=R_1$

$k=1: \cos120^\circ+i\sin120^\circ=-\frac{1}{2}+i\frac{1}{2}\sqrt{3}=R_2$

$k=2: \cos240^\circ+i\sin240^\circ=-\frac{1}{2}-i\frac{1}{2}\sqrt{3}=R_3$

These are the cube roots of 1.

Use De Moivre's Theorem to evaluate $(3 \sqrt3 -3i) ^ 6$ .

Explanation

First convert this point to polar form:

$r = \sqrt{(3\sqrt3)^2 + (-3)^2 } = \sqrt{9 \cdot 3 + 9 } = \sqrt{36} = 6$

$\cos \theta = \frac{ 3\sqrt3}{6 } = \frac{\sqrt3}{2}$

$\theta = \cos ^{-1} (\frac{\sqrt3}{2} ) = \frac{ \pi }{6} \enspace \or \enspace \frac{ 11 \pi }{6}$

Since this number has a negative imaginary part and a positive real part, it is in quadrant IV, so the angle is $\frac{ 11 \pi }{6}$

We are evaluating $(6 \cos \frac{11 \pi }{6} + i \sin \frac{ 11 \pi }{6} ) ^ 6$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get:

$6^6 (\cos \frac{ 11 \pi }{6} \cdot 6 + i \sin \frac{ 11 \pi }{6} \cdot 6)$

$\frac{11 \pi }{6} \cdot 6 = 11 \pi$ which is coterminal with $\pi$ since it is an odd multiplie

$46,656(\cos \pi + i \sin \pi ) = 46,656 (-1 + 0i) = -46,656$

Use De Moivre's Theorem to evaluate .

Explanation

First, convert the complex number to polar form:

$r = \sqrt{ 2^2 + 2^2 } = \sqrt{4 + 4 } = \sqrt{2 \cdot 4 } = 2 \sqrt 2$

$\sin \theta = \frac{ 2 }{ 2 \sqrt 2 } = \frac{ 1 }{ \sqrt2 }$

$\theta = \sin ^{-1 } (\frac{ 1}{ \sqrt 2 }) = \frac{ \pi }{4} \enspace or \enspace \frac{ 3 \pi }{4}$

Since both the real and the imaginary parts are positive, the angle is in quadrant I, so it is $\frac{ \pi }{4}$

This means we're evaluating

$(2 \sqrt2 (\cos \frac{ \pi }{4} + i \sin \frac{ \pi }{4} ))^ 9$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$(2 \sqrt2 )^9 (\cos \frac{\pi }{4} \cdot 9 + i \sin \frac{ \pi }{4} \cdot 9 )$

First, evaluate $(2\sqrt2)^9$ . We can split this into $2^9 \cdot (\sqrt2)^9$ which is equivalent to $2^9 \cdot (\sqrt2)^8 \cdot \sqrt2 = 2^9 \cdot 2^4 \cdot \sqrt 2$

\[We can re-write the middle exponent since $\sqrt 2$ is equivalent to $2^{\frac{1}{2}}$ \]

This comes to $8,192 \sqrt 2$

Evaluating sine and cosine at $\frac{\pi }{4} \cdot 9 = \frac{ 9 \pi }{4}$ is equivalent to evaluating them at $\frac{ \pi }{4}$ since $\frac{ 9 \pi }{4} - 2 \pi = \frac{9 \pi }{4 } - \frac{ 8 \pi }{4} = \frac{ \pi }{4}$