Express the complex number z=1−33iz=1-\frac{\sqrt3}{3}i in polar form.

z=233(cos⁡11π6+isin⁡11π6)z=\frac{2\sqrt3}{3}\left(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}\right)

Express the complex number z = 1 − 3 3 i z=1-\frac{\sqrt3}{3}i in polar form.

Let's see how we can solve this problem. So here we're asked to express the complex number z is equal to 1 minuteus the square root of 3 over 3i in polar form. Now polar form looks something like this, it's gonna be z is equal to r times the cosine of theta, plus I times the sine of theta. Now, r is the square root of x squared plus y squared, where x and y are the real and imaginary parts of our complex number respectively, and theta is going to be the inverse tangent of y over x. So using these two equations, we can plug everything into that equation and get our final answer. Now first, I'm going to start by calculating r. R is going to be the square root of x squared plus y squared. X is the real part which is 1, so we're gonna have one squared plus y which is the imaginary part the imaginary part we can see is negative square root 3 over 3, and then that whole thing's gonna be squared. Now the square root of, well, one square is just gonna be 1, and then for negative square root 3 over 3 squared, I need to square the top and bottom of this fraction. Now the square root of 3 if we square that that's just gonna give us 3 and then we take 3 and square it we're going to get 9 and anytime you score a negative number it will come out to be positive. So this is what we end up with. Now what I can do from here is reduce this fraction. Because 3 over 9 is the same thing as 1 third. Now what I can do is add 1 to 1 third, because one is the same thing as 3 over 3. Because 3 divided by 3 is 1, so we have 3 over 3 plus 1 over 3, and since I have like denominators, I can add the numerators. So we're gonna have 3 plus 1, which is the square root of 4 over 3. Now this whole thing will come out to 2 over the square root of 3, because the square root of 4 is 2, and then the square root of 3 is the square root of 3, then I need to rationalize So So this right here is our solution for R, and that's the R value that we're looking for in the polar form equation. But now we need to figure out what our theta is, And theta, we said is the inverse tangent of y over x. Now I can see here that y is going to be the imaginary part negative square root 3 over 3. And that's gonna be divided by the real part x which is 1. Now because we have 1 in the denominator we can just write this numerator by itself. Going to have the theta is equal to the inverse tangent of just negative square root 3 over 3. Now from here what I can do is plug this into a calculator and see what we get or I can actually take a look at the unit circle because on a unit circle the inverse tangent of square root of 3 over 3 turns out to be 30 degrees or pi over 6 radians and so we have this negative sign and since we're dealing with an inverse tangent this is actually going to come out to negative pi over 6 radians. You can check this on a calculator to be sure because on a calculator, you should get negative 30 degrees, which is equivalent to negative pi over 6 radians. So this right here is our angle theta, but the question becomes, is this the total angle we're looking for? Well, recall that for finding our total angle, it depends on the quadrant we end up in. I can figure out the quadrant by drawing a small graph over here. Now we're going to have a real part of our graph and an imaginary part. And I can take a look at our complex number, and I can see the real part is positive, so so we end up somewhere over here. I can see the imaginary part is negative so that means we're gonna end up somewhere down there on the imaginary axis. So our point actually ends up over here in quadrant 4. Now because we end up in quadrant 4 that means we need to take whatever our angle is and add 360 degrees because what we're ultimately looking for if we have our arc here is the total angle when we start on the positive x axis and go all the way around, meaning we need to add 360 to our angle. Now because we're dealing with radians of this problem, rather than adding 360 degrees, I'm actually going to add 2 pi. This 2 pi is the radian equivalent of 360 degrees. Now what I can do is make sure that we have like denominators, so I can take this 2 pi, and I can multiply this by 6 divided by 6, because that's just the same thing as 2 pi. So if I do this, well, we'll get these like denominators. So we're gonna have that theta is equal to negative pi over 6. 2 times 6 is 12, so we're going to have 12 pi divided by 6. And since I have these like denominators, I can just add the numerators. Negative pi plus 12 pi turns out to be 11 pi. So we have 11 pi over 6 radians as our angle theta. So that's gonna be our angle theta that's the total angle this is the r value. So now that we have both of these we can plug everything into the polar form equation. So we're going to have that z is equal to r and we figured out that r is this value down here it's square root 2 times or 2 times square root 3 over 3. So we're gonna end up with equal to 2 and square root 3 over 3 and that's going to be multiplied by the cosine of our angle which is 11 pi over 6 radians plus I times the sine of 11 pi over 6 radians. And this right here is the solution, this is our complex number in polar form, and that is the final answer. So hope you found this video helpful. Thanks for watching, and please let me know if you have any questions.

Express the complex number z=1−33iz=1-\frac{\sqrt3}{3}i in polar form.

z=233(cos⁡11π6+isin⁡11π6)z=\frac{2\sqrt3}{3}\left(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}\right)

z=233(cos⁡π6−isin⁡π6)z=\frac{2\sqrt3}{3}\left(\cos\frac{\pi}{6}-i\sin\frac{\pi}{6}\right)

z=233(cos⁡13π6+isin⁡13π6)z=\frac{2\sqrt3}{3}\left(\cos\frac{13\pi}{6}+i\sin\frac{13\pi}{6}\right)

z=23(cos⁡11π6+isin⁡11π6)z=\frac{2}{\sqrt3}\left(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}\right)

11. Graphing Complex Numbers

Polar Form of Complex Numbers

Trigonometry

11. Graphing Complex Numbers

Polar Form of Complex Numbers

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Multiple Choice

Express the complex number $z = 1 - \frac{\sqrt{3}}{3} i$ in polar form.

$z = \frac{2 \sqrt{3}}{3} (\cos \frac{π}{6} - i \sin \frac{π}{6})$

$z = \frac{2 \sqrt{3}}{3} (\cos \frac{13 π}{6} + i \sin \frac{13 π}{6})$

$z = \frac{2}{\sqrt{3}} (\cos \frac{11 π}{6} + i \sin \frac{11 π}{6})$

$z = \frac{2 \sqrt{3}}{3} (\cos \frac{11 π}{6} + i \sin \frac{11 π}{6})$

Verified step by step guidance

Identify the given complex number in standard form: z = 1 - $\frac{\sqrt{3}$}{3}i.

Convert the complex number to polar form by finding the magnitude (r) using the formula r = $\sqrt{a^2 + b^2}$, where a is the real part and b is the imaginary part.

Calculate the argument ($\theta$) of the complex number using $\theta$ = $\tan$^{-1}$\left$($\frac{b}{a}$$\right$). Since the complex number is in the fourth quadrant, adjust $\theta$ accordingly.

Express the complex number in polar form as z = r($\cos\[\theta$ + i$\sin\]\theta$). Substitute the calculated values of r and $\theta$.

Verify the polar form by comparing it with the given options and confirm that the correct answer is z = $\frac{2\sqrt{3}$}{3}$\left$($\cos\[\frac{11\pi}{6}$ + i$\sin\]\frac{11\pi}{6}$$\right$).

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Struggling with Trigonometry?

Express the complex number z=1−33iz=1-\(\frac{\sqrt3}{3}\)i in polar form.

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Polar Form of Complex Numbers practice set

Express the complex number $z = 1 - \frac{\sqrt{3}}{3} i$ in polar form.