1. **Graph the complex numbers on the complex plane**. Each complex number $a + bi$ is represented as the point $(a,b)$ where $a$ is the real part (x-axis) and $b$ is the imaginary part (y-axis). - i) $2 + 3i$ corresponds to point $(2,3)$. - ii) $2 - 3i$ corresponds to point $(2,-3)$. - iii) $-2 - 3i$ corresponds to point $(-2,-3)$. - iv) $-2 + 3i$ corresponds to point $(-2,3)$. - v) $-6$ corresponds to point $(-6,0)$. - vi) $i$ corresponds to point $(0,1)$. - vii) $\frac{3}{5} - \frac{4}{5}i$ corresponds to point $(0.6,-0.8)$. - viii) $-5 - 6i$ corresponds to point $(-5,-6)$. 2. **Find the multiplicative inverse of each complex number**. The multiplicative inverse of $z = a + bi$ is $\frac{1}{z} = \frac{\overline{z}}{|z|^2} = \frac{a - bi}{a^2 + b^2}$. - i) For $-3i = 0 - 3i$, inverse is $\frac{0 + 3i}{0^2 + (-3)^2} = \frac{3i}{9} = \frac{1}{3}i$. - ii) For $1 - 2i$, inverse is $\frac{1 + 2i}{1^2 + (-2)^2} = \frac{1 + 2i}{5}$. - iii) For $-3 - 5i$, inverse is $\frac{-3 + 5i}{(-3)^2 + (-5)^2} = \frac{-3 + 5i}{34}$. - iv) For $(1,2)$ interpreted as $1 + 2i$, inverse is $\frac{1 - 2i}{1^2 + 2^2} = \frac{1 - 2i}{5}$. 3. **Simplify powers of $i$**. Recall $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and powers repeat every 4. - i) $i^{101} = i^{(4 \times 25) + 1} = i^1 = i$. - ii) $(-ai)^4 = (-1)^4 a^4 i^4 = a^4 \times 1 = a^4$ since $i^4=1$. - iii) $i^{-3} = \frac{1}{i^3} = \frac{1}{-i} = -\frac{1}{i} = -(-i) = i$ (using $\frac{1}{i} = -i$). - iv) $i^{-10} = \frac{1}{i^{10}} = \frac{1}{i^{8} i^{2}} = \frac{1}{1 \times (-1)} = -1$. 4. **Prove $\overline{z} = z$ iff $z$ is real**. Let $z = a + bi$, then $\overline{z} = a - bi$. If $\overline{z} = z$, then $a - bi = a + bi$ implies $-bi = bi$ so $2bi = 0$ which means $b=0$. Thus $z = a$ is real. 5. **Simplify expressions in form $a + bi$**. - i) $5 + 2\sqrt{-4} = 5 + 2 \times 2i = 5 + 4i$. - ii) $(2 + \sqrt{-3})(3 + \sqrt{-3}) = (2 + i\sqrt{3})(3 + i\sqrt{3})$. Multiply: $2 \times 3 = 6$ $2 \times i\sqrt{3} = 2i\sqrt{3}$ $3 \times i\sqrt{3} = 3i\sqrt{3}$ $i\sqrt{3} \times i\sqrt{3} = i^2 \times 3 = -3$ Sum: $6 + 2i\sqrt{3} + 3i\sqrt{3} - 3 = (6 - 3) + (2i\sqrt{3} + 3i\sqrt{3}) = 3 + 5i\sqrt{3}$. - iii) $\frac{2}{\sqrt{5} + \sqrt{-8}} = \frac{2}{\sqrt{5} + 2i\sqrt{2}}$. Multiply numerator and denominator by conjugate $\sqrt{5} - 2i\sqrt{2}$: Numerator: $2(\sqrt{5} - 2i\sqrt{2}) = 2\sqrt{5} - 4i\sqrt{2}$ Denominator: $(\sqrt{5})^2 - (2i\sqrt{2})^2 = 5 - 4i^2 \times 2 = 5 - 4 \times (-1) \times 2 = 5 + 8 = 13$ Result: $\frac{2\sqrt{5}}{13} - \frac{4\sqrt{2}}{13}i$. - iv) $\frac{3}{\sqrt{6} - \sqrt{-12}} = \frac{3}{\sqrt{6} - 2i\sqrt{3}}$. Multiply numerator and denominator by conjugate $\sqrt{6} + 2i\sqrt{3}$: Numerator: $3(\sqrt{6} + 2i\sqrt{3}) = 3\sqrt{6} + 6i\sqrt{3}$ Denominator: $(\sqrt{6})^2 - (2i\sqrt{3})^2 = 6 - 4i^2 \times 3 = 6 - 4 \times (-1) \times 3 = 6 + 12 = 18$ Result: $\frac{3\sqrt{6}}{18} + \frac{6\sqrt{3}}{18}i = \frac{\sqrt{6}}{6} + \frac{\sqrt{3}}{3}i$. 6. **Show for all $z \in \mathbb{C}$**: - i) $z^2 - \overline{z}^2$ is real. Let $z = a + bi$, then $\overline{z} = a - bi$. Calculate: $z^2 = (a + bi)^2 = a^2 + 2abi + b^2 i^2 = a^2 - b^2 + 2abi$ $\overline{z}^2 = (a - bi)^2 = a^2 - 2abi + b^2 i^2 = a^2 - b^2 - 2abi$ Subtract: $z^2 - \overline{z}^2 = (a^2 - b^2 + 2abi) - (a^2 - b^2 - 2abi) = 4abi$ Since $4abi$ is purely imaginary unless $b=0$, so the statement as is is false unless $b=0$. - ii) $(z - \overline{z})^2$ is real. Calculate: $z - \overline{z} = (a + bi) - (a - bi) = 2bi$ Square: $(2bi)^2 = 4b^2 i^2 = 4b^2 (-1) = -4b^2$ which is real. Hence, $(z - \overline{z})^2$ is always real. **Final answers:** - Multiplicative inverses as above. - Powers of $i$ simplified. - Simplifications in $a + bi$ form. - Proof that $\overline{z} = z$ iff $z$ real. - $(z - \overline{z})^2$ is real; $z^2 - \overline{z}^2$ is purely imaginary unless $b=0$.