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Precalculus : Express Complex Numbers In Polar Form

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Example Questions

Example Question #1 : Express Complex Numbers In Polar Form

The following equation has complex roots:

x^2 - 3x + 9 = 0

Express these roots in polar form.

Possible Answers:

$3(\cos 60 \pm i\sin 60)$

$3(\cos 30 \pm i\sin(30))$

$\sqrt{3}(\cos 60 \pm i\sin 60)$

$\sqrt{3}(\cos 30 \pm i \sin 30)$

Correct answer:

$3(\cos 60 \pm i\sin 60)$

Explanation:

Every complex number can be written in the form a + bi

The polar form of a complex number takes the form r(cos $\Theta$ + isin $\Theta$ )

Now r can be found by applying the Pythagorean Theorem on a and b, or:

r = $\sqrt{a^{2} + b^{2}}$

$\Theta$ can be found using the formula:

$\Theta$ = $tan^{-1}(b/a)$

So for this particular problem, the two roots of the quadratic equation

x^2 - 3x + 9 = 0

are:

$x = \frac{3\pm 3\sqrt{3}i}{2}$

Hence, a = 3/2 and b = 3√3 / 2

Therefore r = $\sqrt{a^{2} + b^{2}}$ = 3

and $\Theta$ = tan^-1 (√3) = 60

And therefore x = r(cos $\Theta$ + isin $\Theta$ ) = 3 (cos 60 + isin 60)

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Example Question #2 : Express Complex Numbers In Polar Form

Express the roots of the following equation in polar form.

x^2+3x+3

Possible Answers:

$\sqrt{3}(\cos(30)+i\sin(30))$

$\sqrt{3}(\cos(150)+i\sin(150))$

$\sqrt{3}(\cos(-30)+i\sin(-30))$

$(\cos(150)+i\sin(150))$

$3(\cos(150)+i\sin(150))$

Correct answer:

$\sqrt{3}(\cos(150)+i\sin(150))$

Explanation:

First, we must use the quadratic formula to calculate the roots in rectangular form.

$\frac{-3\pm \sqrt{9-36}}{2}=\frac{-3\pm i\sqrt{3}}{2}$

Remembering that the complex roots of the equation take on the form a+bi,

we can extract the a and b values.

$a=-\frac{3}{2}$

$b=\frac{\sqrt{3}}{2}$

We can now calculate r and theta.

$r=\sqrt{a^2 + b^2}$

$\theta=\tan^{-1}\frac{b}{a}$

Using these two relations, we get

$r=\sqrt{3}$

$\theta=-30$ . However, we need to adjust this theta to reflect the real location of the vector, which is in the 2nd quadrant (a is negative, b is positive); a represents the x-axis in the real-imaginary plane, b represents the y-axis.

The angle theta now becomes 150.

$\theta=150$ .

You can now plug in r and theta into the standard polar form for a number:

$r(\cos(\theta) +i\sin(\theta ))$

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Example Question #1 : Express Complex Numbers In Polar Form

Express the complex number z=7+11i in polar form.

Possible Answers:

$z=\sqrt{170}(\cos{61.18^\circ +i\sin{61.18^\circ}})$

$z=\sqrt{170}(\cos{45.03^\circ +i\sin{45.03^\circ}})$

$z=77(\cos{45.03^\circ +i\sin{45.03^\circ}})$

$z=13.04(\cos{57.53^\circ +i\sin{57.53^\circ}})$

Correct answer:

$z=13.04(\cos{57.53^\circ +i\sin{57.53^\circ}})$

Explanation:

The figure below shows a complex number plotted on the complex plane. The horizontal axis is the real axis and the vertical axis is the imaginary axis.

Vecc

The polar form of a complex number is $z= r(\cos{\theta}+i\sin{\theta)}$ . We want to find the real and complex components in terms of and $\theta$ where is the length of the vector and $\theta$ is the angle made with the real axis.

We use the Pythagorean Theorem to find :

$r=\sqrt{7^2+11^2}=\sqrt{170} \approx 13.04$

We find $\theta$ by solving the trigonometric ratio

$\tan{\theta} = \frac{11}{7}$

Using $\arctan{\frac{11}{7}} = \theta \approx 57.53$ ,

Then we plug and $\theta$ into our polar equation to obtain

$z=13.04(\cos{57.53^\circ +i\sin{57.53^\circ}})$

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Example Question #4 : Express Complex Numbers In Polar Form

What is the polar form of the complex number $\tiny \tiny \small 3+4i$ ?

Possible Answers:

$\large \large 5e^{\arcsin\frac{4}{3}i}$

$\large \large \large 5e^{\arctan\frac{4}{3}i}$

$\large 25e^{\arcsin\frac{3}{4}i}$

$\large \large \large 25e^{\arctan\frac{4}{3}i}$

$\large \large 5e^{\arctan\frac{3}{4}i}$

Correct answer:

$\large \large \large 5e^{\arctan\frac{4}{3}i}$

Explanation:

The correct answer is

$\small 5e^{\arctan\frac{4}{3}i}$

The polar form of a complex number $\small a+bi$ is $\small re^{i\theta}$ where $\small r$ is the modulus of the complex number and $\small \theta$ is the angle in radians between the real axis and the line that passes through $\small a+bi$ ( $\small r=\sqrt{a^2+b^2}$ and $\small \theta = \arctan(b/a)$ ). We can solve for $\small r$ and $\small \theta$ easily for the complex number $\small 3+4i$ :

$\small r = \sqrt{3^2+4^2}=\sqrt{25}=5$

$\small \theta = \arctan(4/3)$

which gives us

$\small 5e^{\arctan\frac{4}{3}i}$

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Example Question #5 : Express Complex Numbers In Polar Form

Express the complex number in polar form:

4+3i

Possible Answers:

z= 0.64(cos(5)+isin(5))

z= 36.86(cos(.93)+isin(.93))

z= 5(cos(.93)+isin(.93))

z= 25(cos(0.64)+isin(0.64))

z= 5(cos(0.64)+isin(0.64))

Correct answer:

z= 5(cos(0.64)+isin(0.64))

Explanation:

Remember that the standard form of a complex number is: z=a+bi , which can be rewritten in polar form as: $z=r(cos(\theta)+isin(\theta))$ .

To find r, we must find the length of the line 4+3i by using the Pythagorean theorem:

$r=\sqrt{a^2+b^2}$

$r = \sqrt{4^2+3^2}$

$=\sqrt{16+9}$

$= \sqrt{25}$

To find $\theta$ , we can use the equation

$\theta=tan^{-1}\frac{b}{a}$

$=tan^{-1}\frac{3}{4}$

$\approx 0.64$

Note that this value is in radians, NOT degrees.

Thus, the polar form of this equation can be written as

z= 5(cos(0.64)+isin(0.64))

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Example Question #6 : Express Complex Numbers In Polar Form

Express this complex number in polar form.

z=1+i

Possible Answers:

$\left [\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \right]$

$\sqrt{2}\left [1+i\frac{\sqrt{2}}{2} \right]$

$\sqrt{2} \left [1+i \right ]$

None of these answers are correct.

$\sqrt{2}\left [\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \right]$

Correct answer:

$\sqrt{2}\left [\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \right]$

Explanation:

x^2 +y^2=r^2

$x=r \cos\Theta$

$y= r \sin\Theta$

$\tan \Theta = \frac{y}{x}$

Given these identities, first solve for and $\Theta$ . The polar form of a complex number is: $r\left [ \cos \Theta +i \sin \Theta \right ]$

z=1+i

x=1

y=1

$r=\sqrt{2}$

$\tan \Theta = 1$

$\tan \Theta = 1$ at $\frac{\pi}{4}$ (because the original point, (1,1) is in Quadrant 1)

Therefore...

$x=\sqrt{2} \cos(\frac{\pi}{4})=\sqrt{2} \:\frac{\sqrt{2}}{2}$

$y=\sqrt{2} \sin(\frac{\pi}{4})=\sqrt{2} \:\frac{\sqrt{2}}{2}$

$z=\sqrt{2}\left [\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \right]$

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Example Question #501 : Pre Calculus

Convert to polar form: 4 - 3i

Possible Answers:

$5 (\cos 36.87^o + i \sin 36.87^o)$

$\cos 340.529^o + i \sin 340. 529^o$

$5(\cos 143.13^o + i \sin 143.13^o)$

$\cos 75.522^o + i \sin 75.522^o$

$5 (\cos 323.13^o + i \sin 323.13^o)$

Correct answer:

$5 (\cos 323.13^o + i \sin 323.13^o)$

Explanation:

First, find the radius :

$r = \sqrt{4^2 + (-3)^2 } = \sqrt{16 + 9 } = \sqrt{25} = 5$

Then find the angle, thinking of the imaginary part as the height and the radius as the hypotenuse of a right triangle:

$\sin \theta = \frac{ -3}{5}$

$\theta = \sin ^{-1} (-\frac{3}{5}) \approx -36.870$ according to the calculator.

We can get the positive coterminal angle by adding 360 :

-36.87^o + 360^o = 323.13 ^o

The polar form is

$5 (\cos 323.13^o + i \sin 323.13^o)$

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Example Question #3 : Express Complex Numbers In Polar Form

Convert to polar form: 3+2i

Possible Answers:

$\sqrt{13} (\cos 326.31^o + i \sin 326.31^o)$

$\sqrt{13} (\cos 56.31^o + i \sin 56.31^o)$

$\sqrt{13} (\cos 33.69^o + i \sin 33.69^o)$

$\sqrt7 (\cos 130.893^o + i \sin 130.893^o )$

$\sqrt7 (\cos 49.107^o + i \sin 49.107^o)$

Correct answer:

$\sqrt{13} (\cos 33.69^o + i \sin 33.69^o)$

Explanation:

First find the radius, :

$r = \sqrt{3^2 + 2^2 } = \sqrt{9 + 4 } =\sqrt{13}$

Now find the angle, thinking of the imaginary part as the height and the radius as the hypotenuse of a right triangle:

$\sin \theta = \frac{ 2 } {\sqrt{13 }}$

$\theta = \sin ^ {-1} (\frac{ 2 }{\sqrt{13}}) \approx 33.69$ according to the calculator.

This is an appropriate angle to stay with since this number should be in quadrant I.

The complex number in polar form is $\sqrt{13} (\cos 33.69^o + i \sin 33.69 ^o )$

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Example Question #1 : Express Complex Numbers In Polar Form

Convert the complex number -2 + 5i to polar form

Possible Answers:

$\sqrt{29 } (\cos 68.199^o + i \sin 68.199^o )$

$\sqrt{29 } ( \cos 158.199 ^o + i \sin 158.199 ^o )$

$\sqrt{29 } (\cos 248.199^o + i \sin 248.199^o)$

$\sqrt{29} (\cos 111.801 ^o + i \sin 111.801^o )$

$\sqrt{29 } (\cos 291.801^o + i \sin 291.801^o)$

Correct answer:

$\sqrt{29} (\cos 111.801 ^o + i \sin 111.801^o )$

Explanation:

First find :

$r = \sqrt{(-2)^2 + 5^2 } = \sqrt{4 + 25 } = \sqrt{29}$

Now find the angle. Consider the imaginary part to be the height of a right triangle with hypotenuse .

$\sin \theta = \frac{5}{\sqrt{29 } }$

$\theta = \sin ^{-1} (\frac{5}{\sqrt{29}})\approx 68.199 ^o$ according to the calculator.

What the calculator does not know is that this angle is actually located in quadrant II, since the real part is negative and the imaginary part is positive.

To find the angle in quadrant II whose sine is also $\frac{5 }{ \sqrt{29}}$ , subtract from 180 :

180^o - 68.199^o = 111.801^o

The complex number in polar form is $\sqrt{29} (\cos 111.801^o + i \sin 111.801^o )$

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Precalculus : Express Complex Numbers In Polar Form

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