Complex Numbers/Polar Form

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Trigonometry › Complex Numbers/Polar Form

Questions 1 - 10

The polar coordinates $\left ( r,\theta \right )$ of a point are $\left(2.1, \frac{7\pi }{3}\right )$ . Convert these polar coordinates to rectangular coordinates.

Explanation

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = \frac{7\pi }{3}$ :

$x = r \cos \theta = 2.1 \cdot \cos \frac{7\pi }{3} = 2.1 \cdot 0.5 = 1.05$

Likewise, given the polar coordinates $(r,\theta )$ , the -coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = \frac{7\pi }{3}$ :

$y = r \sin \theta = 2.1 \cdot \sin \frac{7\pi }{3} \approx 2.1 \cdot 0.8660 \approx 1.82$

Therefore the rectangular coordinates of the point $\left(2.1, \frac{7\pi }{3}\right )$ are .

The polar coordinates $\left ( r,\theta \right )$ of a point are $\left(2.1, \frac{7\pi }{3}\right )$ . Convert these polar coordinates to rectangular coordinates.

Explanation

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = \frac{7\pi }{3}$ :

$x = r \cos \theta = 2.1 \cdot \cos \frac{7\pi }{3} = 2.1 \cdot 0.5 = 1.05$

Likewise, given the polar coordinates $(r,\theta )$ , the -coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = \frac{7\pi }{3}$ :

$y = r \sin \theta = 2.1 \cdot \sin \frac{7\pi }{3} \approx 2.1 \cdot 0.8660 \approx 1.82$

Therefore the rectangular coordinates of the point $\left(2.1, \frac{7\pi }{3}\right )$ are .

Find the following quotients, given that $z_1=10(\cos40^\circ+i\sin40^\circ)$ and $z_2=2(\cos220^\circ-i\sin220^\circ)$ . Give results in both polar and rectangular forms.

(a) $\frac{z_1}{z_2}$

(b) $\frac{z_2}{z_1}$

(a) $\frac{z_1}{z_2}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_1}{z_2}=-5$

(b) $\frac{z_2}{z_1}=5(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_2}{z_1}=-\frac{1}{5}$

(a) $\frac{z_1}{z_2}=5(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_1}{z_2}=-5$

(b) $\frac{z_2}{z_1}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_2}{z_1}=-\frac{1}{5}$

(a) $\frac{z_1}{z_2}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_1}{z_2}=-5$

(b) $\frac{z_2}{z_1}=5(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_2}{z_1}=-\frac{1}{5}$

(a) $\frac{z_1}{z_2}=5(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_1}{z_2}=-5i$

(b) $\frac{z_2}{z_1}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_2}{z_1}=-\frac{1}{5}i$

Explanation

The modulus of the quotient of two complex numbers is the modulus of the dividend divided by the modulus of the divisor. The amplitude of the quotient is the amplitude of the dividend minus the amplitude of hte divisor.

(a) The modulus for $\frac{z_1}{z_2}$ is equal to $\frac{10}{2}=5$ . The amplitude for $\frac{z_1}{z_2}$ is equal to $40^\circ-220^\circ=-180^\circ=180^\circ$ . (We have chosen to represent this as the coterminal angle $180^\circ$ rather than $-180^\circ$ as it is more conventional to represent angle measures as a positive angle between $0^\circ$ and $360^\circ$ .) Putting this together, we get $\frac{z_1}{z_2}=5(\cos180^\circ+i\sin180^\circ)$ . To represent this in rectangular form, substitute $\cos180^\circ=-1$ and $\sin180^\circ=0$ to get $\frac{z_1}{z_2}=5(-1+0i)=-5$ .

(b) The modulus for $\frac{z_2}{z_1}$ is equal to $\frac{2}{10}=\frac{1}{5}$ . The amplitude for $\frac{z_2}{z_1}$ is equal to $220^\circ-40^\circ=180^\circ$ . Putting this together, we get $\frac{z_2}{z_1}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ . To represent this in rectangular form, substitute $\cos180^\circ=-1$ and $\sin180^\circ=0$ to get $\frac{z_2}{z_1}=\frac{1}{5}(-1+0i)=-\frac{1}{5}$ .

Multiply the following complex numbers (in polar form), giving the result in both polar and rectangular form.

$2(\cos70^\circ+i\sin70^\circ)\cdot 3(\cos20^\circ+i\sin20^\circ)$

$6(\cos90^\circ+i\sin90^\circ)$ or

$5(\cos90^\circ+i\sin90^\circ)$ or

Explanation

The modulus of the product of two complex numbers is the product of their moduli, and the amplitude of the product is the sum of their amplitudes.

Therefore, the new modulus will be $2\cdot 3=6$ and the new amplitude will be $70^\circ+20^\circ=90^\circ$ . Therefore

$2(\cos70^\circ+i\sin70^\circ)\cdot 3(\cos20^\circ+i\sin20^\circ)$ $=6(\cos90^\circ+i\sin90^\circ)$

We must also express this in rectangular form, which we can do by substituting $\cos90^\circ=0$ and $\sin90^\circ=1$ . We get:

$6(\cos90^\circ+i\sin90^\circ)$

Multiply the following complex numbers (in polar form), giving the result in both polar and rectangular form.

$2(\cos70^\circ+i\sin70^\circ)\cdot 3(\cos20^\circ+i\sin20^\circ)$

$6(\cos90^\circ+i\sin90^\circ)$ or

$5(\cos90^\circ+i\sin90^\circ)$ or

Explanation

The modulus of the product of two complex numbers is the product of their moduli, and the amplitude of the product is the sum of their amplitudes.

Therefore, the new modulus will be $2\cdot 3=6$ and the new amplitude will be $70^\circ+20^\circ=90^\circ$ . Therefore

$2(\cos70^\circ+i\sin70^\circ)\cdot 3(\cos20^\circ+i\sin20^\circ)$ $=6(\cos90^\circ+i\sin90^\circ)$

We must also express this in rectangular form, which we can do by substituting $\cos90^\circ=0$ and $\sin90^\circ=1$ . We get:

$6(\cos90^\circ+i\sin90^\circ)$

Express the complex number $z=4(\cos240^\circ+i\sin240^\circ)$ in rectangular form.

$z=-2-2\sqrt{3}i$

$z=2+i\sqrt{3}$

$z=-4-2\sqrt{3}i$

$z=4+i\sqrt{3}$

Explanation

To convert this number to rectangular form, first think about what $\cos240^\circ$ and $\sin240^\circ$ are equal to. Because $240^\circ=180^\circ+60^\circ$ , we can use a 30-60-90o reference triangle in the 3rd quadrant to determine these values.

D5g1377

$\cos240^\circ=-\frac{1}{2}$

$\sin240^\circ=-\frac{\sqrt{3}}{2}$

Now plug these in and continue solving:

$z=4(\cos240^\circ+i\sin240^\circ)$

$z=4(-\frac{1}{2}-i\frac{\sqrt{3}}{2})$

$z=-2-2\sqrt{3}i$