Perform the indicated operations: 1^(°) (1)/(i); 2^(°)…
Perform the indicated operations: $$ 1^{\circ}\,\, \frac{1}{i}; \quad 2^{\circ}\,\, \frac{1-i}{1+i}; \quad 3^{\circ}\,\, \frac{2}{1-3i}. $$
Explanation of Method
To divide by a complex number $c+di$, multiply numerator and denominator by its conjugate $c-di$. Since $(c+di)(c-di)=c^2+d^2$ is real and positive, the denominator becomes real, and the quotient can be written directly in the standard form $a+bi$.
Part $1^{\circ}$ · $\dfrac{1}{i}$
Step 1 · Multiply by the conjugate of $i$
The conjugate of $i$ is $-i$:
$$ \frac{1}{i} = \frac{1}{i}\cdot\frac{-i}{-i} = \frac{-i}{-i^2} = \frac{-i}{-(-1)} = \frac{-i}{1} = -i $$
Step 2 · Quick check via $i^2=-1$
$$ \frac{1}{i} = \frac{1}{i}\cdot\frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i $$
Verification: $i\cdot(-i) = -i^2 = 1$. ✓
Answer for $1^{\circ}$
$$ \boxed{\frac{1}{i} = -i} $$
Part $2^{\circ}$ · $\dfrac{1-i}{1+i}$
Step 1 · Multiply by the conjugate of the denominator
The conjugate of $1+i$ is $1-i$:
$$ \frac{1-i}{1+i} = \frac{1-i}{1+i}\cdot\frac{1-i}{1-i} = \frac{(1-i)^2}{(1+i)(1-i)} $$
Step 2 · Compute numerator and denominator
$$ (1-i)^2 = 1-2i+i^2 = 1-2i-1 = -2i, \qquad (1+i)(1-i) = 1-i^2 = 1+1 = 2 $$
Step 3 · Combine
$$ \frac{-2i}{2} = -i $$
Verification: $(1+i)(-i) = -i-i^2 = -i+1 = 1-i$. ✓
Answer for $2^{\circ}$
$$ \boxed{\frac{1-i}{1+i} = -i} $$
Note this coincides with $1^{\circ}$: $\dfrac{1-i}{1+i}=\dfrac1i=-i$.
Part $3^{\circ}$ · $\dfrac{2}{1-3i}$
Step 1 · Multiply by the conjugate of the denominator
The conjugate of $1-3i$ is $1+3i$:
$$ \frac{2}{1-3i} = \frac{2}{1-3i}\cdot\frac{1+3i}{1+3i} = \frac{2(1+3i)}{(1-3i)(1+3i)} $$
Step 2 · Compute numerator and denominator
$$ 2(1+3i) = 2+6i, \qquad (1-3i)(1+3i) = 1-9i^2 = 1+9 = 10 $$
Step 3 · Combine and simplify
$$ \frac{2+6i}{10} = \frac15+\frac35i $$
Verification: $(1-3i)\left(\dfrac15+\dfrac35i\right) = \dfrac{1-3i+3i+9}{5} = \dfrac{10}{5}=2$. ✓
Answer for $3^{\circ}$
$$ \boxed{\frac{2}{1-3i} = \frac15+\frac35 i = \frac{1+3i}{5}} $$
General method for division
For $z=\dfrac{a+bi}{c+di}$ with $a,b,c,d\in\mathbb{R}$, multiply by $\dfrac{c-di}{c-di}$:
$$ \frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{c^2+d^2} = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i $$
using $i^2=-1$ and $(c+di)(c-di)=c^2+d^2$, always real and non-negative.
| Part | Expression | Result | $|\,\cdot\,|$ |
|---|---|---|---|
| $1^{\circ}$ | $\frac{1}{i}$ | $-i$ | $1$ |
| $2^{\circ}$ | $\frac{1-i}{1+i}$ | $-i$ | $\frac{|1-i|}{|1+i|}=\frac{\sqrt2}{\sqrt2}=1$ |
| $3^{\circ}$ | $\frac{2}{1-3i}$ | $\frac15+\frac35i$ | $\frac{2}{\sqrt{10}}$ |
Result
Every division of complex numbers reduces to multiplying by the conjugate of the denominator, which turns the denominator into the real, positive quantity $c^2+d^2$.