# How do you plot #9sqrt{3} + 9i# on the complex plane and write it in polar form?

So the number in polar coordinates is:

The drawing in the complex plane is a triangle in the first quadrant, with x=9 and y=9sqrt(3).

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To plot (9\sqrt{3} + 9i) on the complex plane, you would locate the point with coordinates ((9\sqrt{3}, 9)).

To express it in polar form, you would first find the magnitude ((r)) and the argument ((\theta)).

The magnitude (r) can be found using the formula (r = \sqrt{a^2 + b^2}), where (a) is the real part and (b) is the imaginary part.

So, (r = \sqrt{(9\sqrt{3})^2 + 9^2} = \sqrt{243 + 81} = \sqrt{324} = 18).

The argument (\theta) can be found using the formula (\theta = \arctan\left(\frac{b}{a}\right)), where (a) and (b) are the real and imaginary parts respectively.

So, (\theta = \arctan\left(\frac{9}{9\sqrt{3}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}).

Therefore, in polar form, (9\sqrt{3} + 9i) can be written as (18(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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