Polar Form of Complex Numbers

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Pre-Calculus › Polar Form of Complex Numbers

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

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

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

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

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

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

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

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

Express the complex number in polar form.

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

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

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

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

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

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

The following equation has complex roots:

Express these roots in polar form.

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

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

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

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

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

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)

Express this complex number in polar form.

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

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

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

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

None of these answers are correct.

Explanation

$x=r \cos\Theta$

$y= r \sin\Theta$

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

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

$r=\sqrt{2}$

$\tan \Theta = 1$

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

Therefore...

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

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

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

Convert to polar form:

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

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

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

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

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

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 :

The polar form is

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

Convert to polar form:

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

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

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

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

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

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

The following equation has complex roots:

Express these roots in polar form.

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

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

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

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

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

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)

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

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

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

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

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

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

Explanation

The correct answer is

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

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

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

which gives us

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

Express this complex number in polar form.

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

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

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

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

None of these answers are correct.

Explanation

$x=r \cos\Theta$

$y= r \sin\Theta$

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

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

$r=\sqrt{2}$

$\tan \Theta = 1$

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

Therefore...

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

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

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

Express the complex number in polar form.

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

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

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

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

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

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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