Complex Numbers – Example

Jumat, 31 Januari 2020

Example 1 Perform the indicated operation and write the answers in standard form.

(a)   (−4 + 7i) + (5 −10i)

(b)   (4 +12i) − (3−15i)

(c)    5i − (−9 + i)

Solution

There really isn’t much to do here other than add or subtract. Note that the parentheses on the first terms are only there to indicate that we’re thinking of that term as a complex number and in general aren’t used.

(a)   (−4 + 7i) + (5 −10i) =1− 3i

(b)   (4 +12i) − (3−15i) = 4 +12i −3+15i =1+ 27i

(c)    5i −(−9 + i) = 5i + 9 − i = 9 + 4i

Example 2 Multiply each of the following and write the answers in standard form.

(a)   7i (−5 + 2i)

(b)   (1−5i)(−9 + 2i)

(c)    (4 + i)(2 + 3i)

(d)   (1−8i)(1+ 8i)

Solution

(a)   So all that we need to do is distribute the 7i through the parenthesis.

7i (−5 + 2i) = −35i +14i²

Now, this is where the small difference mentioned earlier comes into play. This number is NOT in standard form. The standard form for complex numbers does not have an i2 in it. This however is not a problem provided we recall that

i² = −1

Using this we get,

7i (−5 + 2i) = −35i +14(−1) = −14 − 35i

We also rearranged the order so that the real part is listed first.

(b)   In this case we will FOIL the two numbers and we’ll need to also remember to get rid of the i².

(1 − 5i)(−9 + 2i) = −9 + 2i + 45i −10i² = −9 + 47i −10(−1) =1+ 47i

(c)    Same thing with this one.

(4 + i)(2 + 3i) = 8 +12i + 2i + 3i² = 8 +14i + 3(−1) = 5 +14i

(d)   Here’s one final multiplication that will lead us into the next topic.

(1−8i)(1+ 8i) =1+8i −8i − 64i² =1+ 64 = 65

Don’t get excited about it when the product of two complex numbers is a real number. That can and will happen on occasion.

Example 3 Write each of the following in standard form.

Solution

So, in each case we are really looking at the division of two complex numbers. The main idea here however is that we want to write them in standard form. Standard form does not allow for any i's to be in the denominator. So, we need to get the i's out of the denominator.

This is actually fairly simple if we recall that a complex number times its conjugate is a real number. So, if we multiply the numerator and denominator by the conjugate of the denominator we will be able to eliminate the i from the denominator.

Now that we’ve figured out how to do these let’s go ahead and work the problems.

Notice that to officially put the answer in standard form we broke up the fraction into the real and imaginary parts.

(d) This one is a little different from the previous ones since the denominator is a pure imaginary number. It can be done in the same manner as the previous ones, but there is a slightly easier way to do the problem.

First, break up the fraction as follows.

Now, we want the i out of the denominator and since there is only an i in the denominator of the first term we will simply multiply the numerator and denominator of the first term by an i.

Example 4 Multiply the following and write the answer in standard form.

Solution

If we where to multiply this out in its present form we would get,

Now, if we were not being careful we would probably combine the two roots in the final term into one which can’t be done!

So, there is a general rule of thumb in dealing with square roots of negative numbers. When faced with them the first thing that you should always do is convert them to complex number. If we follow this rule we will always get the correct answer.

So, let’s work this problem the way it should be worked.

.

Sumber

Labels: Mathematician

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