Pre-Calculus - Evaluate Powers of Complex Numbers Using DeMoivre's Theorem

Find the magnitude of :

${}(1+3i)^{7}$ , where the complex number satisfies ${}i^{2}=-1$ .

${}10^3\sqrt{10}$

${}10^7\sqrt{10}$

${}10\sqrt{10}$

${}10^{14}\sqrt{10}$

${}10^2\sqrt{10}$

Explanation

Note for any complex number z, we have:

${}|z^n|=|z|^n$ .

Let . Hence ${}|z|=\sqrt{10}$

Therefore:

${}|z^{7}|=|z|^{7}=(\sqrt{10})^{7}=(\sqrt{10})^{6}(\sqrt{10})$

This gives the result.

What is the magnitude of ?

$\sqrt13$

Explanation

To find the magnitude of a complex number we use the following formula:

$|z|=\sqrt{a^2+b^2}$ , where .

Therefore we get,

$|z|=\sqrt{2^2+3^2}=\sqrt{13}$ .

Now to find

$|z|^2=(\sqrt{13})^2$

Evaluate:

Explanation

First, convert this complex number to polar form.

$r= \sqrt{ a^2 + b^2 } = \sqrt{1^2 +(-1)^2 }=\sqrt{2}$

$\sin \theta = \frac{-1}{\sqrt2}$

$\theta = \frac{5 \pi }{4} \enspace or \enspace \frac{7 \pi }{4}$

Since the point has a positive real part and a negative imaginary part, it is located in quadrant IV, so the angle is $\frac{7\pi}{4}$ .

This gives us $(\sqrt{2} ( \cos \frac{7 \pi }{4} + i \sin \frac{ 7 \pi }{4} )) ^ 8$

To evaluate, use 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.

$(\sqrt2 )^8 ( \cos \frac{ 7 \pi }{4} \cdot 8 + i \sin \frac{ 7 \pi }{4} \cdot 8 )$ simplifying

$2 ^ 4 ( \cos 14 \pi + i \sin 14 \pi )$ , $14 \pi$ is coterminal with since it is an even multiple of $\pi$

$2^ 4 (\cos 0 + i \sin 0 ) = 16(1 + 0i) = 16$

Evaluate $(2 -2 \sqrt3)^6$

Explanation

First, convert this complex number to polar form:

$r = \sqrt{2^2 + (2\sqrt3)^2 } = \sqrt{4 +12} = \sqrt{16} = 4$

$\sin \theta = \frac{ -2\sqrt3 }{4} = -\frac{\sqrt3}{2}$

$\theta = \sin ^{-1} (-\frac{\sqrt3}{2}) = \frac{4 \pi }{3} \enspace or \enspace \frac{ 5 \pi }{3}$

Since the real part is positive and the imaginary part is negative, this is in quadrant IV, so the angle is $\frac{5 \pi }{3}$

So we are evaluating $(4(\cos \frac{5 \pi }{3} + i \sin \frac{5 \pi }{3} ) )^6$

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.

$4^6 (\cos \frac{ 5 \pi }{3} \cdot 6 + i \sin \frac{ 5 \pi }{3} \cdot 6 ) = 4,096 ( \cos 10 \pi + i \sin 10 \pi)$

$10 \pi$ is coterminal with since it is an even multiple of $\pi$

$4,096( \cos 0 + i \sin 0 ) = 4,096 ( 1 + 0i) = 4,096$

Simplify

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

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

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

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

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

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.

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

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

Use DeMoivre's Theorem to evaluate the expression $(\sqrt3 - 3i ) ^ 6$ .

Explanation

First convert this complex number to polar form:

$r = \sqrt{ (\sqrt3)^2 + 1^2 } = \sqrt{ 3 + 1 } = \sqrt4 = 2$

$\sin \theta = \frac{ 1}{ 2}$ so $\theta = \frac{\pi }{6} \enspace or \enspace \frac{5 \pi }{6}$

Since this number has positive real and imaginary parts, it is in quadrant I, so the angle is $\frac{ \pi }{6}$

So we are evaluating $(2(\cos \frac{ \pi }{6} + i \sin \frac{ \pi }{6} ))^3$

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.

$2^3 (\cos \frac{3\pi}{6} + i \sin \frac{ 3\pi }{6} ) = 8 ( \cos \frac{\pi}{2} + i \sin \frac{ \pi }{2} ) = 8(0 + i ) = 8i$

$(3 \sqrt3 -3i) ^ 6$

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

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

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 $\frac{ 9 \pi }{4} - 2 \pi = \frac{9 \pi }{4 } - \frac{ 8 \pi }{4} = \frac{ \pi }{4}$