\(\displaystyle x + y-2z = (4-2i) + (6+7i) -2(5i) = 10 - 5i = 5(2-i)\)

When dealing with imaginary numbers, we multiply by foiling as we do with binomials. When we do this we get the expression below: 

\(\displaystyle (3+i)(4+i)= 12 + 3i + 4i + i^2\)

Since we know that \(\displaystyle i^2 = -1\) we get \(\displaystyle 12 + 7i - 1\) which gives us \(\displaystyle 11 + 7i\). 

Recall that the definition of imaginary numbers gives that \(\displaystyle i = \sqrt{-1}\) and thus that \(\displaystyle i^2 = -1\). Therefore, we can use Exponent Rules to write \(\displaystyle i^4 = i^2 \cdot i^2 = -1\cdot-1 = 1\)

When adding complex numbers, add the real parts and the imaginary parts separately to get another complex number in standard form.

Adding the real parts gives \(\displaystyle 2-1=1\), and adding the imaginary parts gives \(\displaystyle -5i\).

Multiply both the numerator and the denominator by the conjugate of the denominator which is \(\displaystyle 3-2i\) which results in

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

The numerator after simplification give us \(\displaystyle 6-4i-3i+2i^{2}=6-7i+2i^{2}=4-7i\)

The denominator is equal to \(\displaystyle 3^{2}-4i^{2}=9+4=13\)

Hence, the final answer in standard form =

\(\displaystyle \frac{4}{13}-\frac{7i}{13}\)

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.

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

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

Combine like terms:

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

Distribute:

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

Combine like terms:

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

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

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

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

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

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

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

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 \(\displaystyle i^2=-1\)

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

\(\displaystyle 19-7i\)