Question

If $\frac{3}{2 + \cos \theta + i \sin \theta} = a + ib$, then prove that $a^2 + b^2 = 4a - 3$.

Answer 1

a^2 + b^2 = 4a - 3

Answer 2

The problem asks us to prove a relationship between $a$, $b$, and $\theta$ given a complex number equation.

Let the given equation be: $\frac{3}{2 + \cos \theta + i \sin \theta} = a + ib$ Let $z = \cos \theta + i \sin \theta$. We know that $z$ is a complex number on the unit circle, so its modulus $|z| = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1$. An important property for complex numbers on the unit circle is that $\bar{z} = \frac{1}{z}$.

The given equation can be written as: $a + ib = \frac{3}{2 + z}$

We need to prove $a^2 + b^2 = 4a - 3$.

Step 1: Express $a^2 + b^2$ in terms of $\theta$. We know that $a^2 + b^2 = |a + ib|^2$. $a^2 + b^2 = \left| \frac{3}{2 + z} \right|^2 = \frac{|3|^2}{|2 + z|^2} = \frac{9}{|2 + z|^2}$ Now, let's find $|2 + z|^2$. For any complex number $w$, $|w|^2 = w \bar{w}$. $|2 + z|^2 = (2 + z)(2 + \bar{z})$ Since $|z|=1$, we have $\bar{z} = \frac{1}{z}$. Substitute this into the expression: $|2 + z|^2 = (2 + z)\left(2 + \frac{1}{z}\right) = 4 + \frac{2}{z} + 2z + 1 = 5 + 2\left(z + \frac{1}{z}\right)$ Now substitute $z = \cos \theta + i \sin \theta$ and $\frac{1}{z} = \cos \theta - i \sin \theta$: $z + \frac{1}{z} = (\cos \theta + i \sin \theta) + (\cos \theta - i \sin \theta) = 2 \cos \theta$ So, $|2 + z|^2 = 5 + 2(2 \cos \theta) = 5 + 4 \cos \theta$ Therefore, $a^2 + b^2 = \frac{9}{5 + 4 \cos \theta} \quad \cdots (1)$

Step 2: Express $a$ in terms of $\theta$. From $a + ib = \frac{3}{2 + z}$, we can find $a$ by taking the real part: $a = \text{Re}\left( \frac{3}{2 + z} \right)$ To find the real part, we multiply the numerator and denominator by the conjugate of the denominator: $a = \text{Re}\left( \frac{3(2 + \bar{z})}{(2 + z)(2 + \bar{z})} \right) = \text{Re}\left( \frac{3(2 + \bar{z})}{|2 + z|^2} \right)$ Substitute $|2 + z|^2 = 5 + 4 \cos \theta$ and $\bar{z} = \cos \theta - i \sin \theta$: $a = \text{Re}\left( \frac{3(2 + \cos \theta - i \sin \theta)}{5 + 4 \cos \theta} \right)$ The real part is: $a = \frac{3(2 + \cos \theta)}{5 + 4 \cos \theta}$

Step 3: Express $4a - 3$ in terms of $\theta$. Substitute the expression for $a$ into $4a - 3$: $4a - 3 = 4\left( \frac{3(2 + \cos \theta)}{5 + 4 \cos \theta} \right) - 3$ $4a - 3 = \frac{12(2 + \cos \theta)}{5 + 4 \cos \theta} - 3$ To combine these terms, find a common denominator: $4a - 3 = \frac{12(2 + \cos \theta) - 3(5 + 4 \cos \theta)}{5 + 4 \cos \theta}$ $4a - 3 = \frac{24 + 12 \cos \theta - 15 - 12 \cos \theta}{5 + 4 \cos \theta}$ $4a - 3 = \frac{9}{5 + 4 \cos \theta} \quad \cdots (2)$

Step 4: Compare the results. From (1) and (2), we have: $a^2 + b^2 = \frac{9}{5 + 4 \cos \theta}$ $4a - 3 = \frac{9}{5 + 4 \cos \theta}$ Since both expressions are equal to $\frac{9}{5 + 4 \cos \theta}$, we can conclude: $a^2 + b^2 = 4a - 3$ Hence, the proof is complete.