Find the real and imaginary parts of the numbers: 1^(°)…
Find the real and imaginary parts of the numbers: $$ 1^{\circ}\,\, \frac{3i}{i-2}; \quad 2^{\circ}\,\, \frac{1+i}{2-i}(3+2i); \quad 3^{\circ}\,\, \frac{1-3i}{i+1}-\frac{i}{i+2}; \quad 4^{\circ}\,\, \frac{\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}}{3-4i}; \quad 5^{\circ}\,\, \frac{i}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}. $$
Explanation of Method
Each expression is simplified by multiplying every fraction by the conjugate of its denominator, turning the denominator into a real number $c^2+d^2$. Trigonometric numerators are first evaluated at their known angles, then treated the same way. Once everything is combined into the standard form $a+bi$, the real and imaginary parts are read directly: $\text{Re}=a$, $\text{Im}=b$.
Part $1^{\circ}$ · $\dfrac{3i}{i-2}$
Step 1 · Multiply by the conjugate
Write the denominator as $-2+i$; its conjugate is $-2-i$.
$$ \frac{3i}{-2+i}\cdot\frac{-2-i}{-2-i} = \frac{3i(-2-i)}{(-2+i)(-2-i)} $$
Step 2 · Compute numerator and denominator
$$ 3i(-2-i) = -6i-3i^2 = 3-6i, \qquad (-2+i)(-2-i) = 4-i^2 = 5 $$
Step 3 · Simplify
$$ \frac{3-6i}{5} = \frac{3}{5}-\frac{6}{5}i $$
Answer for $1^{\circ}$
$$ \boxed{\text{Re}=\frac35,\quad \text{Im}=-\frac65} $$
Part $2^{\circ}$ · $\dfrac{1+i}{2-i}(3+2i)$
Step 1 · Simplify $\dfrac{1+i}{2-i}$
Multiply by the conjugate $2+i$:
$$ \frac{(1+i)(2+i)}{(2-i)(2+i)} = \frac{2+i+2i+i^2}{4-i^2} = \frac{1+3i}{5} $$
Step 2 · Multiply by $(3+2i)$
$$ \frac{1+3i}{5}\cdot(3+2i) = \frac{(1+3i)(3+2i)}{5} $$
Step 3 · Expand and simplify
$$ (1+3i)(3+2i) = 3+2i+9i+6i^2 = -3+11i \quad\Longrightarrow\quad \frac{-3+11i}{5} = -\frac35+\frac{11}{5}i $$
Answer for $2^{\circ}$
$$ \boxed{\text{Re}=-\frac35,\quad \text{Im}=\frac{11}{5}} $$
Part $3^{\circ}$ · $\dfrac{1-3i}{i+1}-\dfrac{i}{i+2}$
Step 1 · Simplify $\dfrac{1-3i}{1+i}$
Multiply by the conjugate $1-i$:
$$ \frac{(1-3i)(1-i)}{(1+i)(1-i)} = \frac{1-i-3i+3i^2}{1-i^2} = \frac{-2-4i}{2} = -1-2i $$
Step 2 · Simplify $\dfrac{i}{2+i}$
Multiply by the conjugate $2-i$:
$$ \frac{i(2-i)}{(2+i)(2-i)} = \frac{2i-i^2}{4-i^2} = \frac{1+2i}{5} $$
Step 3 · Subtract
$$ (-1-2i)-\frac{1+2i}{5} = \left(-1-\frac15\right)+i\left(-2-\frac25\right) = -\frac65-\frac{12}{5}i $$
Answer for $3^{\circ}$
$$ \boxed{\text{Re}=-\frac65,\quad \text{Im}=-\frac{12}{5}} $$
Part $4^{\circ}$ · $\dfrac{\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}}{3-4i}$
Step 1 · Evaluate the trigonometric numerator
$$ \cos\frac{\pi}{6}=\frac{\sqrt3}{2},\quad \sin\frac{\pi}{6}=\frac12 \quad\Longrightarrow\quad \cos\frac{\pi}{6}+i\sin\frac{\pi}{6} = \frac{\sqrt3+i}{2} $$
Step 2 · Multiply by the conjugate of $3-4i$
$$ \frac{\sqrt3+i}{2(3-4i)}\cdot\frac{3+4i}{3+4i} = \frac{(\sqrt3+i)(3+4i)}{2(3-4i)(3+4i)} $$
Step 3 · Compute numerator and denominator
$$ (\sqrt3+i)(3+4i) = 3\sqrt3+4\sqrt3i+3i+4i^2 = (3\sqrt3-4)+(4\sqrt3+3)i $$
$$ (3-4i)(3+4i)=9-16i^2=25 \quad\Longrightarrow\quad 2\cdot25=50 $$
Step 4 · Simplify
$$ \frac{(3\sqrt3-4)+(4\sqrt3+3)i}{50} = \frac{3\sqrt3-4}{50}+\frac{4\sqrt3+3}{50}i $$
Answer for $4^{\circ}$
$$ \boxed{\text{Re}=\frac{3\sqrt3-4}{50}\approx0.024,\quad \text{Im}=\frac{4\sqrt3+3}{50}\approx0.199} $$
Part $5^{\circ}$ · $\dfrac{i}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}$
Step 1 · Evaluate the trigonometric denominator
$$ \cos\frac{\pi}{3}=\frac12,\quad \sin\frac{\pi}{3}=\frac{\sqrt3}{2} \quad\Longrightarrow\quad \cos\frac{\pi}{3}+i\sin\frac{\pi}{3} = \frac{1+\sqrt3i}{2} $$
Step 2 · Set up the division
$$ \frac{i}{\frac{1+\sqrt3i}{2}} = \frac{2i}{1+\sqrt3i} $$
Step 3 · Multiply by the conjugate
$$ \frac{2i(1-\sqrt3i)}{(1+\sqrt3i)(1-\sqrt3i)} = \frac{2\sqrt3+2i}{1+3} = \frac{2\sqrt3+2i}{4} = \frac{\sqrt3}{2}+\frac12 i $$
Answer for $5^{\circ}$
$$ \boxed{\text{Re}=\frac{\sqrt3}{2},\quad \text{Im}=\frac12} $$
Check via Euler's formula
Since $\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=e^{i\pi/3}$ and $i=e^{i\pi/2}$:
$$ \frac{i}{e^{i\pi/3}} = e^{i(\pi/2-\pi/3)} = e^{i\pi/6} = \cos\frac{\pi}{6}+i\sin\frac{\pi}{6} = \frac{\sqrt3}{2}+\frac12 i $$
which confirms the result of Step 3.
| Part | Real part | Imaginary part |
|---|---|---|
| $1^{\circ}$ | $\frac35$ | $-\frac65$ |
| $2^{\circ}$ | $-\frac35$ | $\frac{11}{5}$ |
| $3^{\circ}$ | $-\frac65$ | $-\frac{12}{5}$ |
| $4^{\circ}$ | $\frac{3\sqrt3-4}{50}$ | $\frac{4\sqrt3+3}{50}$ |
| $5^{\circ}$ | $\frac{\sqrt3}{2}$ | $\frac12$ |
Result
General method: multiply every fraction by the conjugate of its denominator, combine terms, then read $\text{Re}=a$, $\text{Im}=b$ from the resulting standard form $a+bi$.