-3 plus 2i-11-14

Answer 1

You would have to convert into the form a + bi where a are real numbers which are not attached to i and bi, a number which is.

Reading from left to right and following the order of operations of PEMDAS (parentheses, exponents, multiplication and division, addition and subtraction), separate the constants and the imaginary number which includes the i:

real numbers, a: -3, -11, -14 & imaginary numbers, b: +2i

then add/subtract values, remembering that negative signs and subtraction signs can be used interchangeably:

-3 - 11 - 14 = -28; thus a which refers to the final/total value of real numbers is -28

...and ...

+2i is the only imaginary number, meaning it is the only value on the right, bi side,making this a + bi complex number:

-28 + 2i

Take into consideration that a real number can have negative or positive value, be a decimal, a fraction, be under a radical, or even be 0. Also regard that imaginary numbers can be all of these as long as the symbol of "i" is added. Imaginary numbers can also be treated similar to real numbers; depending on the signs and operations specified, the imaginary numbers can be added/subtracted by doing the operation(s) on its coefficient(s) and "is" when multiplying/dividing (since you are distributing):

Adding:

(7 + 8i) + (6 + 4i) = 13 + 12i

Subtracting:

(8 - 5i) + (6 - 3i) = 14 - 8i

Multiplying:

(4 + 3i) * (5 - 2i) = (4 * 5) + (4 * -2i) + (5 * 3i) + (3i * -2i) = 20 + -8i + 15i -6i^2 = 20 -6(-1) -8i + 15i = 26 + 7i

Dividing:

(8 + 8i) / (2 + 2i) = (8 + (8 * i)) / (2 + (2 * i)) = (8+8)(i)/(2+2)(i) =16/4 = 4 ("i"s cancel)

Moreover, remember that i = the square root of -1 but when i is squared, i^2, the square root symbol and squared sign cancel, making the value of i^2, -1, a real number as the "i" is no longer present.