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Express the given complex number in the form \(R(\cos \theta+\) \(i \sin \theta)=R e^{i \theta}\) $$ -i $$

Short Answer

Expert verified
Answer: The polar and exponential form for the complex number -i is 1e^(i(3π/2)).

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01

Find the magnitude (R)

Using the polar representation of complex numbers, R is the magnitude of the complex number. For the complex number, which is given as \(-i\), we can write it as \(0-1i.\) Now, use the formula \(R=\sqrt{a^2+b^2}\) where a is the real part, and b is the imaginary part. In our case, a = 0 and b = -1. So, $$R=\sqrt{(0)^2+(-1)^2} = \sqrt{0+1} = \sqrt{1} =1.$$
02

Find the angle (θ)

To find the angle (θ), we can use the formula \(\theta=\tan^{-1}\frac{b}{a}\), where a is the real part and b is the imaginary part. Keep in mind that we need to consider which quadrant the complex number lies in. Since our complex number is \(-i\), it is in the third quadrant (negative real axis and negative imaginary axis). In our case, a = 0 and b = -1. So, $$\theta=\tan^{-1}\frac{-1}{0}.$$ Since the division by zero is undefined and we know that the complex number is on the negative imaginary axis, we can determine that $$\theta=\frac{3\pi}{2}$$ using the knowledge of the coordinates system and trigonometry.
03

Write the complex number in the desired form

Now that we have the magnitude (R) and the angle (θ), we can write the complex number in the required form: $$R(\cos\theta+\( \)i\sin\theta)=R e^{i\theta}.$$ For our case, R = 1 and θ = \(\frac{3\pi}{2}\). So, $$1(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}) = 1e^{i\frac{3\pi}{2}}.$$ Thus, the desired form for the complex number \(-i\) is: $$-i = 1e^{i\frac{3\pi}{2}}.$$

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Representation
Complex numbers can be understood more intuitively using polar representation, especially when dealing with multiplication, division, and roots. Polar representation expresses a complex number in terms of its magnitude and angle from the positive real axis.

A complex number like \(-i\) can be expressed as \(R (\cos \theta + i \sin \theta) = R e^{i \theta}\), where:
  • \(R\) is the magnitude of the complex number (think of it as the distance from the origin to the point in the complex plane).
  • \(\theta\) is the angle, measured in radians, from the positive real axis to the line connecting the origin and the point representing the complex number.
Getting comfortable with this representation helps simplify calculations involving complex arithmetic. For example, when multiplying two complex numbers, you multiply their magnitudes and add their angles, which is quite straightforward in polar form. This is why polar representation is such a powerful tool.
Magnitude of Complex Number
The magnitude of a complex number represents its absolute value or "size" in the complex plane. It is symbolized by \(R\) and calculated using the Pythagorean theorem.

For a complex number represented as \(a + bi\), the formula for magnitude is:
  • \(R = \sqrt{a^2 + b^2}\)
In the case of \(-i\), which can be written as \(0 - 1i\), the real part (\(a\)) is 0, and the imaginary part (\(b\)) is -1. Plugging these values into the formula gives:
  • \(R = \sqrt{0^2 + (-1)^2} = \sqrt{0 + 1} = \sqrt{1} = 1\)
The magnitude tells us that the complex number \(-i\) is exactly 1 unit away from the origin on the complex plane. Understanding magnitude is critical, as it allows for comparison between different complex numbers and plays a crucial role in operations like division.
Angle of Complex Number
The angle of a complex number, often denoted as \(\theta\), describes its direction in the complex plane relative to the positive real axis. It is typically determined using the inverse tangent function, \(\tan^{-1}\), sometimes producing undefined results when the real part is zero.

For \(-i\), we have the real part \(a = 0\) and imaginary part \(b = -1\). Usually, \(\theta\) would be computed using:
  • \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\)
Since \(a = 0\) leads to division by zero, one must consider the position of the complex number. \(-i\) lies on the negative imaginary axis, placing it in the third quadrant of the complex plane.

Thus, knowing the geometry and using trigonometry, we determine:
  • \(\theta = \frac{3\pi}{2}\) radians
This corrects for the typical ambiguity in calculating angle when dealing with specific axis-aligned complex numbers. The angle combined with the magnitude completes the polar representation, essential for various complex number operations.

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Most popular questions from this chapter

Find the general solution of the given differential equation. $$ 12 y^{\mathrm{iv}}+31 y^{\prime \prime \prime}+75 y^{\prime \prime}+37 y^{\prime}+5 y=0 $$

We consider another way of arriving at the proper form of \(Y(t)\) for use in the method of undetermined coefficients. The procedure is based on the observation that exponential, polynomial, or sinusoidal terms (or sums and products of such terms) can be viewed as solutions of certain linear homogeneous differential equations with constant coefficients. It is convenient to use the symbol \(D\) for \(d / d t\). Then, for example, \(e^{-t}\) is a solution of \((D+1) y=0 ;\) the differential operator \(D+1\) is said to annihilate, or to be an annihilator of, \(e^{-t}\). Similarly, \(D^{2}+4\) is an annihilator of \(\sin 2 t\) or \(\cos 2 t,\) \((D-3)^{2}=D^{2}-6 D+9\) is an annihilator of \(e^{3 t}\) or \(t e^{3 t},\) and so forth. Show that linear differential operators with constant coefficients obey the commutative law, that is, $$ (D-a)(D-b) f=(D-b)(D-a) f $$ for any twice differentiable function \(f\) and any constants \(a\) and \(b .\) The result extends at once to any finite number of factors.

Follow the procedure illustrated in Example 4 to determine the indicated roots of the given complex number. $$ [2(\cos \pi / 3+i \sin \pi / 3)]^{1 / 2} $$

Use the method of variation of parameters to determine the general solution of the given differential equation. $$ y^{\prime \prime \prime}+y^{\prime}=\sec t, \quad-\pi / 2

Find the solution of the given initial value problem. Then plot a graph of the solution. \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=3 t+4, \quad y(0)=y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=y^{\prime \prime \prime}(0)=1\)

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