Multiply both the numerator and the denominator by the conjugate of the denominator which is \(\displaystyle -i\) resulting in

\(\displaystyle \frac{\left ( 2+3i \right )\left ( -i \right )}{\left ( i \right )\left ( -i \right )}\)

This is equal to \(\displaystyle \frac{-2i-{3i}^2}{-i^2}\)

Since \(\displaystyle i^2 = -1\) you can make that substitution of \(\displaystyle -1\) in place of \(\displaystyle i^2\) in both numerator and denominator, leaving:

\(\displaystyle \frac{-2i-3(-1)}{-(-1)}\)

When you then cancel the negatives in both numerator and denominator (remember that \(\displaystyle -(-1)=1\), simplifying each term), you're left with a denominator of \(\displaystyle 1\) and a numerator of \(\displaystyle -2i + 3\), which equals \(\displaystyle 3-2i\).

Use the FOIL method to simplify. FOIL means to mulitply the first terms together, then multiply the outer terms together, then multiply the inner terms togethers, and lastly, mulitply the last terms together.

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

\(\displaystyle =i[(1)(2)+(1)(-i)+(i)(2)+(i)(-i)]\)

\(\displaystyle =i[2-i+2i-i^2]\)

\(\displaystyle =i[2+i-i^2]\)

\(\displaystyle =2i+i^2-i^3\)

The imaginary \(\displaystyle i\) is equal to:

\(\displaystyle \sqrt{-1}\)

Write the terms for \(\displaystyle i, i^2, \textup{and } i^3\).

\(\displaystyle i=\sqrt{-1}\)

\(\displaystyle i^2=-1\)

\(\displaystyle i^3=i \times i^2=-\sqrt{-1}=-i\)

Replace \(\displaystyle i^2 \textup{ and } i^3\) with the appropiate values and simplify.

\(\displaystyle 2i+i^2-i^3= 2i-1-(-i) = -1+3i\)

We know that \(\displaystyle \small i=\sqrt{-1}}\). Therefore, \(\displaystyle \small i^{2}=-1, i^{3}=-\sqrt{-1}, i^{4}=1\). Thus, every exponent of \(\displaystyle \small i\) that is a multiple of 4 will yield the value of \(\displaystyle \small 1\). This makes \(\displaystyle \small i^{40} =1\). Since \(\displaystyle \small i^{42}=i^{40}*i*i\), we know that \(\displaystyle \small i^{42}=1*i*i=i^{2}=-1\).

\(\displaystyle (3+2i)+4-i-i(7+i)\)

Combine like terms:

\(\displaystyle 7+i-i(7+i)\)

Distribute:

\(\displaystyle 7+i-7i-i^2\)

Combine like terms:

\(\displaystyle 7+i-7i-(-1)\)

\(\displaystyle \mathbf{8-6i}\)

\(\displaystyle \sqrt{-4}*\sqrt{-25}*\sqrt{-64}\)

\(\displaystyle 2i*5i*8i=80i^3\)

\(\displaystyle i^3=-1\)

\(\displaystyle \mathbf{-80i}\)

To rationalize a complex fraction, multiply numerator and denominator by the conjugate of the denominator.

\(\displaystyle \frac{(2+i)(3+i)}{(3-i)(3+i)}\)

\(\displaystyle \frac{6+3i+2i+i^2}{9-3i+3i-i^2}\)

\(\displaystyle \frac{5+5i}{10}\)

\(\displaystyle \frac{1+i}{2}\)

To rationalize a complex fraction, multiply numerator and denominator by the conjugate of the denominator.

\(\displaystyle \frac{(2+i)(3+i)}{(3-i)(3+i)}\)

\(\displaystyle \frac{6+3i+2i+i^2}{9-3i+3i-i^2}\)

\(\displaystyle \frac{5+5i}{10}\)

\(\displaystyle \frac{1+i}{2}\)

Distribute:

\(\displaystyle 3i(3-4i) = 9i + 12\)

\(\displaystyle 1(3-4i) = 3 - 4i\)

combine like terms:

\(\displaystyle 5i + 15\)

Use FOIL to multiply the two binomials.

Recall that FOIL stands for Firsts, Outers, Inners, and Lasts.

\(\displaystyle (3+i)(5-4i)\)

\(\displaystyle 15+5i-12i-4i^2\)

Remember that \(\displaystyle i^2=-1\)

\(\displaystyle 15-7i+4\)

\(\displaystyle 19-7i\)

Distribute the minus sign:

\(\displaystyle 5-3i-2-2i\)

Combine like terms:

\(\displaystyle 3-5i\)