Ask a question
0 0

complex numbers

perform each of the following calculations involving complex numbers, and write the answer, simplified as much as possible, in the form a + bi, where a and b are real numbers.
1.  (4+2i) – (6 – 8i) (work not required)
2.  i18 (work not required)
3.  (1-3i)/(2+ i) Show work.
Tutors, please sign in to answer this question.

1 Answer

1.  (4+2i) - (6-8i):
distribute the minus sign:
= 4 + 2i - 6 + 8i
gather like terms:
= -2 + 10i
2. i18 = i times 18.  You should write it as 0 + 18i to give the answer in the form asked for.
3. (1-3i) / (2+i):
here you have to multiply the top AND bottom by the "complex conjugate" of the bottom - much in the same way you simplify radical expressions to get rid of radicals in the denominator.
the complex conjugate of any complex number, a + bi, is a - bi
So you multiply the given expression by  (2-i ) / (2 - i)
The numerator becomes (1 - 3i)(2 - i) which you can foil, or use the distribute law, or the "claw method" - whatever is easiest for you since these are all really the same method:
(1 - 3i)(2 - i) = 2 - 6i - i + 3i^2 =  2 - 7i +3(-1) = 2 - 7i - 3 = -1 - 7i
(remember i^2 = -1 )
The denominator becomes (2+i)(2-i) = 4 + 2i - 2i - i^2 = 4 - (-1) = 4 + 1 = 5
ANSWER:  (-1 - 7i) / 5 which can be written in a + bi form as
- 1/5 - (7/5) i 


i^18 = -1
There is an interesting repeating pattern when it comes to powers of i.
any number to the 0 power = 1, so i^0 = 1
any number to the 1st power is that number, so i ^ 1 = i
i^2 is -1 since (sqrt (-1)) squared = -1.  (Remember this, because you use this to substitute on the next 2 powers.)
i^3 is i times i^2 = i times (-1) = -i.
I usually go one more to make it easy upon myself to figure out what the higher powers of i are:
i^4 = i^2 times i^2 = (-1)(-1) = 1.  Just like i^0 = 1.

This is easy to remember because you already know the first 2 in this series (i^0 and i^1) and the next two or three make sense or are easy to figure out.
So now I have
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
The pattern repeats so
i^5 = i^1 = i
i^6 =i^2 = -1
i^7 =i^3 = -i
i^8 =i^4 = 1
and so on.
so i^18 I would figure like this:
I could keep making a chart of the powers of i.  But there is a faster way once you see the pattern.  Make your chart for up to i^4.  Then find the multiple of 4 closest, but less than the power.  In this case 16 is the largest multiple of 4 less than 18.  So i^16 = i^4 =   1.  The count two more in the sequence.  i^17 would be like i^1 = i.  i^18 would be like i^2, or -1.
Would be so much easier to show you, than write it out in words, but basically you are just counting around the pattern up to the power you are looking for.  
An alternate and elegant way of doing this would be to set up the complex coordinate axes (x is normal, but replace y with i).  i^0 is the unit vector on x, pointing in the positive direction, which =1.  i^1 is like rotating the unit vector 90 degrees, so now it is pointing up one unit in the positive direction on the complex (or i, (like the y)) axis, so it is i.  i^2 is rotating it 90 more degrees so it is the unit vector pointing to -1 on x.  i^3 is rotating the unit vector 90 more degrees, so now it is pointing down towards -i because it is on the complex (or i, again, like the y) axis.  One more 90 degree rotation, which is a full rotation, you are back to the unit vector point to +1 on the x axis.  Again the picture is hard to describe in words, but if you get it in your head, it makes these sorts of problems really easy.  Going to larger powers means you are making more rotations, but the pattern repeats.
This is easy to remember because you already know the first 2 in this series (i^0 and i^1) and the next two make sense or are easy to figure out.