Express Complex Numbers In Polar Form - Pre-Calculus

Card 0 of 9

Question

The following equation has complex roots:

Express these roots in polar form.

Answer

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)

Compare your answer with the correct one above

Question

Express the roots of the following equation in polar form.

Answer

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:

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

Compare your answer with the correct one above

Question

Express the complex number in polar form.

Answer

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.

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

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

Express this complex number in polar form.

Answer

$x=r \cos\Theta$

$y= r \sin\Theta$

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

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...

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

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

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

Compare your answer with the correct one above

Question

Express the complex number in polar form:

Answer

Remember that the standard form of a complex number is: , 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 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

Compare your answer with the correct one above

Question

Convert to polar form:

Answer

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

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

Compare your answer with the correct one above

Question

Convert to polar form:

Answer

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

Compare your answer with the correct one above

Question

Convert the complex number to polar form

Answer

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

Compare your answer with the correct one above

Tap the card to reveal the answer