Math problems on complex numbers and conjugates.

Here are the step-by-step solutions for each problem in the list.

1. Express $(1+3i)^{-1}$ in the form $a + ib$

To write this in standard form, we need to move the imaginary unit out of the denominator. We do this by multiplying the numerator and denominator by the conjugate of the denominator, which is $1 - 3i$.

$$
(1+3i)^{-1} = \frac{1}{1+3i} \times \frac{1-3i}{1-3i}
$$

The denominator becomes a difference of squares: $(1)^2 - (3i)^2 = 1 - 9(-1) = 1 + 9 = 10$.
The numerator is simply $1 - 3i$.

$$
= \frac{1-3i}{10} = \frac{1}{10} - \frac{3}{10}i
$$

Answer: $\frac{1}{10} - \frac{3}{10}i$

---

2. Explain the fallacy in $-1 = i \cdot i = \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1$

The error occurs at the step $\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)}$.
The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is only valid when $a$ and $b$ are non-negative real numbers. When dealing with negative numbers or complex numbers, this property does not hold because the square root function is multi-valued in the complex plane. Specifically, defining $\sqrt{-1}$ as $i$ fixes a specific branch, but multiplying them requires careful handling of arguments (angles), leading to $i^2 = -1$, not $\sqrt{1}$.

Answer: The identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$ fails for negative real numbers.

---

3. Find the conjugate of $\frac{1}{2-3i}$

First, simplify the fraction into $a+ib$ form by rationalizing the denominator. Multiply top and bottom by the conjugate $2+3i$:

$$
\frac{1}{2-3i} \times \frac{2+3i}{2+3i} = \frac{2+3i}{4 - 9(-1)} = \frac{2+3i}{13} = \frac{2}{13} + \frac{3}{13}i
$$

The conjugate changes the sign of the imaginary part:

$$
\overline{\left(\frac{2}{13} + \frac{3}{13}i\right)} = \frac{2}{13} - \frac{3}{13}i
$$

Answer: $\frac{2}{13} - \frac{3}{13}i$

---

4. Find the conjugate of $-3i - 5$

Rewrite the number in standard $a+ib$ form: $-5 - 3i$.
The conjugate is found by changing the sign of the imaginary part ($-3i$ becomes $+3i$).

Answer: $-5 + 3i$

---

5. Let $z_1 = 2-i, z_2 = -2+i$. Find $\text{Re}\left(\frac{z_1 z_2}{z_1}\right)$

Notice that the expression simplifies before doing complex arithmetic. Since $z_1 \neq 0$, we can cancel $z_1$ from the numerator and denominator:

$$
\frac{z_1 z_2}{z_1} = z_2
$$

So we just need the real part of $z_2$. Given $z_2 = -2 + i$, the real part is $-2$.

Answer: $-2$

---

6. If $x - iy = \sqrt{\frac{a-ib}{c-id}}$, prove that $(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$

Take the modulus (absolute value) of both sides. Recall that $|x-iy| = \sqrt{x^2+y^2}$ and $|\sqrt{w}| = \sqrt{|w|}$.

$$
|x-iy| = \left| \sqrt{\frac{a-ib}{c-id}} \right|
$$

$$
\sqrt{x^2+y^2} = \sqrt{ \left| \frac{a-ib}{c-id} \right| } = \sqrt{ \frac{|a-ib|}{|c-id|} } = \sqrt{ \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} }
$$

Square both sides to remove the outer square root:
$$
x^2+y^2 = \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}
$$

Square both sides again to remove the remaining square roots:
$$
(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}
$$

Answer: Proven.

---

7. If $a+ib = \frac{c+i}{c-i}$, prove $a^2+b^2=1$ and $\frac{b}{a} = \frac{2c}{c^2-1}$

Part 1: Prove $a^2+b^2=1$
$a^2+b^2$ is the square of the modulus $|a+ib|^2$.
$$
|a+ib| = \left| \frac{c+i}{c-i} \right| = \frac{|c+i|}{|c-i|} = \frac{\sqrt{c^2+1}}{\sqrt{c^2+(-1)^2}} = \frac{\sqrt{c^2+1}}{\sqrt{c^2+1}} = 1
$$
Since the modulus is 1, $a^2+b^2 = 1^2 = 1$.

Part 2: Prove $\frac{b}{a} = \frac{2c}{c^2-1}$
Rationalize the RHS to find $a$ and $b$:
$$
a+ib = \frac{c+i}{c-i} \times \frac{c+i}{c+i} = \frac{c^2 + 2ci + i^2}{c^2+1} = \frac{(c^2-1) + i(2c)}{c^2+1}
$$
Equating real and imaginary parts:
$a = \frac{c^2-1}{c^2+1}$ and $b = \frac{2c}{c^2+1}$.

Now divide $b$ by $a$:
$$
\frac{b}{a} = \frac{2c/(c^2+1)}{(c^2-1)/(c^2+1)} = \frac{2c}{c^2-1}
$$

Answer: Proven.

---

8. If $z_1 = 2+i$ and $z_2 = 1+i$, find $\left| \frac{z_1+z_2+1}{z_1-z_2+i} \right|$

Calculate the numerator and denominator separately.
Numerator: $z_1 + z_2 + 1 = (2+i) + (1+i) + 1 = 4 + 2i$.
Denominator: $z_1 - z_2 + i = (2+i) - (1+i) + i = 2 + i - 1 - i + i = 1 + i$.

Now find the modulus of the fraction:
$$
\left| \frac{4+2i}{1+i} \right| = \frac{|4+2i|}{|1+i|} = \frac{\sqrt{4^2+2^2}}{\sqrt{1^2+1^2}} = \frac{\sqrt{20}}{\sqrt{2}} = \sqrt{10}
$$

Answer: $\sqrt{10}$

---

9. If $(p+iq)^2 = x+iy$, prove $(p^2+q^2)^2 = x^2+y^2$

Expand the left side: $(p+iq)^2 = p^2 - q^2 + 2ipq$.
So, $x = p^2-q^2$ and $y = 2pq$.

Calculate $x^2+y^2$:
$$
x^2+y^2 = (p^2-q^2)^2 + (2pq)^2 = p^4 - 2p^2q^2 + q^4 + 4p^2q^2 = p^4 + 2p^2q^2 + q^4
$$
This factors perfectly into $(p^2+q^2)^2$.

Alternatively, using moduli: $|(p+iq)^2| = |x+iy| \implies |p+iq|^2 = \sqrt{x^2+y^2}$. Squaring both sides gives $(p^2+q^2)^2 = x^2+y^2$.

Answer: Proven.

---

10. Convert $\frac{-16}{1+i\sqrt{3}}$ into polar form

First, simplify the complex number to rectangular form by rationalizing:
$$
\frac{-16}{1+i\sqrt{3}} \times \frac{1-i\sqrt{3}}{1-i\sqrt{3}} = \frac{-16(1-i\sqrt{3})}{1+3} = \frac{-16+16i\sqrt{3}}{4} = -4 + 4i\sqrt{3}
$$

Find the modulus $r$:
$r = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8$.

Find the argument $\theta$:
$\tan \alpha = |\frac{4\sqrt{3}}{-4}| = \sqrt{3} \implies \alpha = \frac{\pi}{3}$.
Since the real part is negative and imaginary part is positive, the point is in Quadrant II.
$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

Polar form is $r(\cos \theta + i \sin \theta)$.

Answer: $8\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right)$