Question - Complex Numbers in Polar Form
Solution:
To write a complex number in polar form, we need to calculate its magnitude (r) and the angle (theta, θ) it makes with the positive real axis.The polar form of a complex number is given by: $$ r(\cos(\theta) + i\sin(\theta)) $$For $$ w = 2 - 2i $$:a) To find the magnitude (r), use the formula $$ r = \sqrt{a^2 + b^2} $$, where a is the real part and b is the imaginary part. So:\[ r_w = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \]The angle (θ) is given by $$ \theta = \arctan\left(\frac{b}{a}\right) $$, however, since the complex number is in the fourth quadrant, we need to add $$ 2\pi $$ to the result to get the positive angle.\[ \theta_w = \arctan\left(\frac{-2}{2}\right) = \arctan(-1) \]In radians, $$ \arctan(-1) $$ corresponds to $$ -\frac{\pi}{4} $$, but we want the positive angle, so we add $$ 2\pi $$:\[ \theta_w = -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4} \]Polar form of $$ w $$ would then be:\[ w = \sqrt{8}\left(\cos\left(\frac{7\pi}{4}\right) + i\sin\left(\frac{7\pi}{4}\right)\right) \]For $$ z = \frac{\sqrt{3}}{2} + \frac{1}{2}i $$:b) The magnitude (r) is:\[ r_z = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]The angle (θ) for $$ z $$ is in the first quadrant, and the real and imaginary parts correspond to the sine and cosine of $$ \frac{\pi}{6} $$, so:\[ \theta_z = \frac{\pi}{6} \]Polar form of $$ z $$ would then be:\[ z = 1\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) \]For $$ z^3 $$, since $$ |z| = 1 $$ and $$ \theta = \frac{\pi}{6} $$:c) Raising $$ z $$ to the third power multiplies its angle by 3, and its magnitude to the power of 3. Since the magnitude of $$ z $$ is 1, it stays 1. Thus:\[ z^3 = 1^3\left(\cos\left(3\cdot\frac{\pi}{6}\right) + i\sin\left(3\cdot\frac{\pi}{6}\right)\right) \]\[ z^3 = 1\left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right) \]Note that $$ \cos\left(\frac{\pi}{2}\right) $$ is 0, and $$ \sin\left(\frac{\pi}{2}\right) $$ is 1, so $$ z^3 $$ can be further simplified to:\[ z^3 = i \]For part d), we are not asked for the exponential form, but in case you need help in the future:d) The exponential form of a complex number is given by $$ r \, e^{i\theta} $$. Using Euler's formula $$ e^{i\theta} = \cos(\theta) + i\sin(\theta) $$, we get:\[ w = \sqrt{8} \, e^{i\frac{7\pi}{4}} \]
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com