For example, to simplify the sum of (a+bi) and (c+di), first identify that a and c are the real number portions, and add them together. Symbolically, this will be (a+c). Using actual numbers instead of variables, consider the example of (3+3i) + (5-2i). The real portion of the first number is 3, and the real portion of the second complex number is 5. Add these together to get 3+5=8. The real portion of the simplified complex number will be 8.
For the algebraic example of (a+bi) plus (c+di), the imaginary portions are b and d. Adding these together algebraically gives the result (b+d)i. Using the numerical example of (3+3i) + (5-2i), the imaginary portions of the two complex numbers are 3i and -2i. Adding these gives the result of 1i, which can also be written just as i.
The sum of (a+bi) and (c+di) is written as (a+c) + (b+d)i. Applying the numerical example, the sum of (3+3i) + (5-2i) is 8+i.
First. The F in FOIL means that you multiply the first term of the first binomial by the first term of the second binomial. For the sample, this would be ac. Outer. The O in FOIL tells you to multiply the “outer” terms. These are the first term of the first binomial and the second term of the second binomial. For the sample, this would be ad. Inner. The I in FOIL means to multiply the “inner” terms. These would be the two terms that appear in the middle, which are the second term of the first binomial and the first term of the second binomial. In the given example, the inner terms are bc. Last. The L in FOIL represents the last terms of each binomial. For the sample expression, this would be bd. Finally, add all four products together. The result for the sample binomial multiplication of (a+b)(c+d) is ac+ad+bc+bd.
First. The product of the first terms is 35=15. Outer. The product of the outer terms is 3(-3i). This product is -9i. Inner. The product of the two inner terms is 2i5. This product is 10i. Last. The product of the last terms is (2i)(-3i). This product is -6i2. Recognize that i2 equals -1, so the value of -6i2 is -6*-1, which is 6.
For the sample 15-9i+10i+6, you can add the 15 and 6 together and add the -9i and the 10i together. The result will be 21+i.
(3)(-2)=-6 (First) (3)(-5i)=-15i (Outer) (4i)(-2)=-8i (Inner) (4i)(-5i)=-20i2=(-20)(-1)=20 (Lasts) -6-15i-8i+20 = 14-23i (Combine terms and simplify)
(4+3i)(2−2i){\displaystyle {\frac {(4+3i)}{(2-2i)}}}
(4+3i)(2−2i)∗(2+2i)(2+2i){\displaystyle {\frac {(4+3i)}{(2-2i)}}{\frac {(2+2i)}{(2+2i)}}} Then multiply the numerator and denominator and simplify as follows: (4+3i)(2−2i)∗(2+2i)(2+2i){\displaystyle {\frac {(4+3i)}{(2-2i)}}{\frac {(2+2i)}{(2+2i)}}} 8+8i+6i+6i24+4i−4i−4i2{\displaystyle {\frac {8+8i+6i+6i^{2}}{4+4i-4i-4i^{2}}}} 8+14i+6(−1)4−4(−1){\displaystyle {\frac {8+14i+6(-1)}{4-4(-1)}}} 8+14i−64+4{\displaystyle {\frac {8+14i-6}{4+4}}} 2+14i8{\displaystyle {\frac {2+14i}{8}}} Notice in the second step above, the denominator contains the terms +4i{\displaystyle +4i} and −4i{\displaystyle -4i}. These will cancel each other out. This will always happen as a result of multiplying by the conjugate. The imaginary terms of the denominator should always cancel and disappear.
2+14i8=28+14i8=14+7i4{\displaystyle {\frac {2+14i}{8}}={\frac {2}{8}}+{\frac {14i}{8}}={\frac {1}{4}}+{\frac {7i}{4}}}
