Step 1: We will multiply the two complex conjugates: \(\displaystyle 2x-3\) and \(\displaystyle 2x+3\).

\(\displaystyle (2-3i)(2+3i)=4+6i-6i-9i^2=4-9i^2\)

Step 2: Replace \(\displaystyle i^2\) with \(\displaystyle -1\).

\(\displaystyle 4-9(-1)\)

Simplify:

\(\displaystyle 4+9(1)\)
\(\displaystyle =13\)


Step 3: Multiply the result of the complex conjugates to the other parentheses,\(\displaystyle (4-3i)\).

\(\displaystyle 13(4-3i)=52-39i\)

The final answer after the product of all three binomials is \(\displaystyle 52-39i\)
 

Quick Way:
Step 1: Expand \(\displaystyle (i-1)^2\) .

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

Remember: \(\displaystyle i^2=-1\)

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

Step 2: \(\displaystyle (1-i)^{10}=(i-1)^{2\cdot 5}\)

By this equivalence, I can just raise the answer of \(\displaystyle (1-i)^2\) to the power \(\displaystyle 5\).

\(\displaystyle (i-1)^{10}=(-2i)^{5}=-32i^5\)

\(\displaystyle i^5 \equiv i^1=i\). Replace \(\displaystyle i^5\)..

Final answer: \(\displaystyle -32i\)

Long Way:

Step 1: FOIL:

Recall, FOIL means to multiply the first terms in both binomials together, the outer terms together, the inner terms together, and finally, the last terms together.

\(\displaystyle (2+3i)(3-4i)=6-8i+9i-12i^2\)

Step 2: Simplify:

\(\displaystyle 6+i-12i^2\)

Step 3: Recall: \(\displaystyle i^2=-1\). Replace and simplify.

\(\displaystyle 6+i-12(-1)\)
\(\displaystyle 6+i+12\)
\(\displaystyle =18+i\)

When adding imaginary numbers, simply add the real parts and the imaginary parts. 

\(\displaystyle (3-5i)+(2+7i)= (3+2)+(-5i+7i)=5+2i\)

\(\displaystyle When\ subtracting\ binomials\ in\ ( )\; first\ distribute\ the\ negative\ sign.\)

\(\displaystyle (13-7i)-(10-8i)=13-7i-10+8i\)

\(\displaystyle Now\ combine\ like\ terms.\)

\(\displaystyle 13-10-7i+8i=3+i\)

Distribute and Multiply:

\(\displaystyle (3-4i)(2i+3)=6i+9-8i^2-12i\)

Simplify all terms...

\(\displaystyle 6i+9-8(-1)-12i\)

\(\displaystyle =6i+9+8-12i\)

\(\displaystyle =-6i+17\)

Step 1: Recall the cycle of imaginary numbers to a random power \(\displaystyle x\).

If \(\displaystyle x=1\), then \(\displaystyle i^1=i\)

If \(\displaystyle x=2\), then \(\displaystyle i^2=-1\)

If \(\displaystyle x=3\), then \(\displaystyle i^3=-i\)

If \(\displaystyle x=4\), then \(\displaystyle i^4=1\)

If \(\displaystyle x=5\), then \(\displaystyle i^5=i^4(i^1)=1(i)=i\)

and so on....

The cycle repeats every \(\displaystyle 4\) terms. 

For ANY number \(\displaystyle x\ge 5\), you can break down that term into smaller elementary powers of i. 

Step 2: Distribute the \(\displaystyle i^{24}\) to all terms in the parentheses:

\(\displaystyle (i^{24})(i^3)+(i^{24})(i^{-2})\).

Step 3: Recall the rules for exponents:

\(\displaystyle \forall\{a,b,c,d\} \in \mathbb{R}::\)

\(\displaystyle (a^b)\cdot (a^c)=a^{b+c}\)

\(\displaystyle {a^{-b}}=\frac {1}{a^b}\)

\(\displaystyle (a^b)\cdot (a^{-c})=a^b \cdot \bigg (\frac {1}{a^c}\bigg)=\frac {a^b}{a^c}=a^{b-c}=a^d\)

Step 4: Use the rules to rewrite the expression in Step 2:

\(\displaystyle (i^{24})\cdot {i^3}=i^{24+3}=i^{27}\)

\(\displaystyle (i^{24})(i^{-2})=\frac {i^{24}}{i^2}=i^{24-2}=i^{22}\)

Step 5: Simplify the results in Step 4. Use the rules in Step 1.:

\(\displaystyle i^{27}=(i^{24}\cdot i^3)=1(-i)=-i\)

\(\displaystyle i^{22}=(i^{20}\cdot i^{2})=1(-1)=-1\)

Step 6: Write the answer in \(\displaystyle a+bi\) form, where \(\displaystyle a\) is the real part and \(\displaystyle bi\) is the imaginary part:

We get \(\displaystyle -1-i\)

When adding complex numbers, we add the real numbers and add the imaginary numbers. 

\(\displaystyle Add: (3-5i)+(2+7i)\)

\(\displaystyle =(3+2)+(-5i+7i)\)

\(\displaystyle =5+2i\)

In order to subtract complex numbers, we must first distribute the negative sign to the second complex number. 

\(\displaystyle Subtract: (15+7i)-(12+9i)\)

\(\displaystyle 15+7i-12-9i\)

\(\displaystyle 15-12+7i-9i=3-2i\)

\(\displaystyle Simplify: 3i(-5+2i)\)

First we must distribute

\(\displaystyle 3i(-5+2i)=-15i+2i^2\)

\(\displaystyle We\ must\ remember\ that\ i^2=-1\)

\(\displaystyle -15i+6i^2=-15i+6(-1)\)

\(\displaystyle -15i+6(-1)=-6-15i\ (i\ term\ is\ always\ last)\)