Question

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

Short answer

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

Step-by-step solution

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, $$R=\sqrt{(0)^2+(-1)^2} = \sqrt{0+1} = \sqrt{1} =1.$$

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.

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: $$-i = 1e^{i\frac{3\pi}{2}}.$$