Multiply both numberator and denominator by $\displaystyle i$:

$\displaystyle \frac{9+4i}{4i}$

$\displaystyle = \frac{\left (9+4i \right )\cdot i}{4i \cdot i}$

$\displaystyle = \frac{9i + 4i^{2}}{4 i^{2}}$

$\displaystyle = \frac{9i + 4 (-1)}{4 (-1)}$

$\displaystyle = \frac{-4+9i }{-4 }$

$\displaystyle = \frac{-4 }{-4 }+ \frac{ 9i }{-4 }$

$\displaystyle = 1- \frac{ 9 }{4 } i$

First, divide 100 by $\displaystyle 5i$ as follows:

$\displaystyle \frac{100}{5i} = \frac{100 \cdot i}{5i \cdot i} = \frac{100 i}{5(-1)} = \frac{100 i}{-5}= -20i$

Now dvide this result by $\displaystyle 5i$:

$\displaystyle \frac{-20i}{5i} = \frac{-20}{5} = -4$

First, divide 100 by $\displaystyle 1+i$ as follows:

$\displaystyle \frac{100}{1+i}$

$\displaystyle =\frac{100(1-i)}{\left (1+i \right )(1-i)}$

$\displaystyle = \frac{100-100i}{1^{2}+1^{2}}$

$\displaystyle = \frac{100-100i}{2}$

$\displaystyle = 50-50i$

Now, divide this by $\displaystyle 1-i$:

$\displaystyle \frac{ 50-50i}{1-i} = \frac{ 50 (1-i)}{1-i} = 50$

First, evaluate $\displaystyle (4i)^{2}$:

$\displaystyle \left (4i \right )^{2} = 16 i^{2} = 16 (-1)= -16$

Now divide this into $\displaystyle 100 i$:

$\displaystyle \frac{100 i}{-16} =- \frac{25}{4}i$

First, evaluate $\displaystyle \left ( 1+ i\right )^{2}$ using the square pattern:

$\displaystyle \left ( 1+ i\right )^{2}$

$\displaystyle =1^{2}+ 2 \cdot 1 \cdot i+ i^{2}$

$\displaystyle =1+2i+ (-1)$

$\displaystyle =2i$

Divide this into $\displaystyle i$:

$\displaystyle \frac{i}{2i} = \frac{1}{2}$

This problem can be solved very similarly to a binomial such as $\displaystyle a+bx$. In this case, both the real and nonreal terms in the complex number are eligible to be divided by the real divisor.

$\displaystyle \frac{6+12i}{4} = (\frac{6}{4} + \frac{12i}{4})$

$\displaystyle \frac{12i}{4} = \frac{12}{4} \cdot i = 3 \cdot i = 3i$, so

$\displaystyle \frac{3}{2} + 3i$

Solving this problem using a conjugate is just like conjugating a binomial to simplify a denominator.

$\displaystyle \frac{(4+2i)(-3+2i)}{(-3-2i)(-3+2i)}$ Multiply both terms by the denominator's conjugate.

$\displaystyle \frac{(4+2i)(-3+2i)}{(-3-2i)(-3+2i)} = \frac{(4+2i)(-3+2i)}{13}$ Simplify. Note $\displaystyle i^2 = -1$.

$\displaystyle (4+2i)(-3+2i) = (-12 + 8i -6i -4)$ Combine and simplify.

$\displaystyle (-12-2i-4) = -16-2i$ Simplify the numerator.

$\displaystyle \frac{-16-2i}{13}$ The prime denominator prevents further simplifying.

Thus, $\displaystyle \frac{4+2i}{-3-2i} = \frac{-16-2i}{13}$.

This problem can be solved in a way similar to other kinds of division problems (with binomials, for example). We need to get the imaginary number out of the denominator, so we will multiply the denominator by its conjugate and multiply the top by it as well to preserve the number's value.

$\displaystyle \frac{5+7i}{2+i} \cdot \frac{2-i}{2-i}=\frac{10-5i+14i-7i^2}{4-i^2}{}$

Then, recall $\displaystyle i^2 = -1$ by definition, so we can simplify this further:

$\displaystyle \frac{10+7+14i-5i}{4+1} = \frac{17}{5}+\frac{9}{5}i{}$

This is as far as we can simplify, so it is our final answer.