Pre-Calculus - Express Complex Numbers In Polar Form

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

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

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 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

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

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

Convert the complex number to polar form

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

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

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

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

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

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 :

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

Express the roots of the following equation in polar form.

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

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

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

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

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

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.