### All SAT Mathematics Resources

## Example Questions

### Example Question #1 : Working With Imaginary Numbers

Simplify:

**Possible Answers:**

**Correct answer:**

It can be easier to line real and imaginary parts vertically to keep things organized, but in essence, combine like terms (where 'like' here means real or imaginary):

### Example Question #2 : Working With Imaginary Numbers

Which of the following complex numbers is equal to ? (Note: )

**Possible Answers:**

**Correct answer:**

When adding and subtracting complex numbers, the “” functions just like a regular variable, the same as if it were “” or any other letter variable. It is only when multiplying and dividing complex numbers that there is a special step where is transformed into . This question simply asks you to subtract one complex number from another one, so “” functions just like any other letter variable.

First subtract the real number parts: . Then subtract the imaginary number parts: . Putting the parts together gives you the **correct answer choice** .

The tricky step is . Subtracting a negative is the same thing as adding a positive since the negative signs cancel each other out. Failure to do this step correctly can lead to the wrong answer choice .

Students who are expecting the question to be more challenging than it actually is may see the complex numbers in parentheses and jump to the conclusion that they need to multiply the complex numbers rather than just subtract them. Incorrectly multiplying the complex numbers rather than subtracting them will lead to the wrong answer choice .

Multiplying the complex numbers and also making a mistake when multiplying may lead to the wrong answer choice .

### Example Question #3 : Working With Imaginary Numbers

Which of the following complex numbers is equal to , where ?

**Possible Answers:**

**Correct answer:**

When adding and subtracting complex numbers, the “” functions just like a regular variable, the same as if it were “” or any other letter variable. It is only when multiplying and dividing complex numbers that there is the special step where is transformed into . This question simply asks you to subtract one complex number from another one, so “” functions just like any other letter variable.

First subtract the real number parts: . Then subtract the imaginary number parts: . Putting the parts together gives you the correct answer choice .

To do both these subtraction steps correctly, the student needs to keep track of all the signs accurately and understand that subtracting a negative is the same thing as adding a positive, since the negative signs cancel each other out. Failure to do this step correctly with the imaginary number parts can lead to the wrong answer choice . Failure to do this step correctly with the real number parts and the imaginary number parts can lead to the wrong answer choice .

Students who are expecting the question to be more challenging than it actually is may see the complex numbers in parentheses and jump to the conclusion that they need to multiply the complex numbers rather than just subtract them. Incorrectly multiplying the complex numbers rather than subtracting them will lead to the wrong answer choice .

### Example Question #4 : Working With Imaginary Numbers

What is the sum ? (Note: )

**Possible Answers:**

**Correct answer:**

When adding and subtracting complex numbers, the “” functions just like a regular variable, the same as if it were “” or any other letter variable. It is only when multiplying and dividing complex numbers that there is the special step where is transformed into . This question simply asks you to add two complex numbers, so “” functions just like any other letter variable.

First add the real number parts . Then add the imaginary number parts . Putting the parts together gives you the correct answer choice .

It is easy to get confused when adding two negative numbers, which is the other main challenge in this question. Many students see it as a subtraction operation, so they want to say that . But when both numbers are negative, it works more like addition, but with every number having a negative sign. Failure to understand this concept and perform this operation correctly with the real number parts may lead to the wrong answer choice . Failure to understand this concept and perform this operation correctly with the imaginary number parts may lead to the wrong answer choice Failure to understand this concept and perform this operation correctly with both parts may lead to the wrong answer choice .

### Example Question #1 : Exponents & Roots

Which of the following complex numbers is equal to , for ?

**Possible Answers:**

**Correct answer:**

There are a few tricks to this question. First of all, you must carefully observe the minus sign in the middle of the expression, which means that you are subtracting the complex numbers, not multiplying them!

Second, you must note the term in the second complex number. Now you have to know that because , therefore . So . Now our whole expression is .

Finally, now you have to be very careful about all the minus signs! For both of the terms in the second complex number, you are now subtracting a negative, which means that you are actually just doing the same thing as adding a positive! Therefore, the whole expression is simply equivalent to . Now it looks much easier, doesn’t it? Simply adding together the real number parts and the imaginary number parts separately, you get the final answer, the **correct answer choice** .

If you do not simplify correctly, or if you accidentally subtract instead of add in the final step, you may get one of the wrong answer choices or . If you accidentally subtract instead of add in the final step, you may get one of the wrong answer choices or .

### Example Question #6 : Working With Imaginary Numbers

Which of the following complex numbers is equal to , for ?

**Possible Answers:**

**Correct answer:**

There are a few tricks to this question. First of all, you must carefully observe the minus sign in the middle of the expression, which means that you are subtracting the complex numbers, not multiplying them!

Second, you must note the term in the second complex number. Now you have to know that because , therefore . So . Now our whole expression is .

Finally, now you have to be very careful about all the minus signs! For both of the terms in the second complex number, you are now subtracting a negative, which means that you are actually just doing the same thing as adding a positive! Therefore, the whole expression is simply equivalent to . Now it looks much easier, doesn’t it? Simply adding together the real number parts and the imaginary number parts separately, you get the final answer, the correct answer choice .

If you do not simplify correctly, or if you accidentally subtract instead of add in the final step, you may mistakenly think the “” and “” terms cancel out, and get one of the wrong answer choices or . If you accidentally subtract instead of add in the final step, you may get one of the wrong answer choices 8i or .

### Example Question #7 : Working With Imaginary Numbers

Which of the following complex numbers is equivalent to ? (Note: )

**Possible Answers:**

**Correct answer:**

This is the classic more difficult type of complex number/imaginary number question on the SAT math section: Division of one complex number by another complex number. The most difficult and critical key step of the solution process is the very first step: You must multiply both the numerator and denominator by what we call the complex conjugate of the denominator, which is a fancy term that just means you change the sign in the middle of the complex number (before the imaginary part). The reason we do this is that this will *make the imaginary part in the denominator disappear!* Recall the difference of squares formula in algebra: . This formula works because when you FOIL , the two middle terms will cancel each other out. Well, the same thing happens here when you multiply the complex number in the denominator by its complex conjugate: the imaginary number middle terms will cancel each other out! So . The middle terms cancel each other out, leaving us with . Now you have to know that because , therefore So , and now the negative signs in the second term cancel out, making it positive: . The point of this whole process is that the end result is to simplify the denominator to a single real number, which in this case is .

To finish solving the question, now you must also multiply the original numerator by the complex conjugate of the denominator: . The two middle terms, the imaginary parts, combine: .

**Pro Tip:** If you are alert here, you can already see that the imaginary part of the final answer will have in the numerator and in the denominator! This alone eliminates all of the wrong answer choices, so you can already see now that the **correct answer choice** must be !

If you do need to finish the solution and know the numerator of the real part as well, you continue , since you must know that as we stated above. Finishing the simplification of the numerator, we get . Now we see how the entire **correct answer choice** is right.

You may wonder how experienced math students can finish solving questions so fast and complete both SAT math sections within the strict time limits and still get perfect 800 SAT Math scores. The answer is, they use the four answer choices to guide them and speed up the process dramatically. Look at the answer choices and . Do you see how they have exactly the same numbers and coefficients as the numerator and denominator of the expression in the original question? Well, experienced math students know that simplification of a fraction like this one with two or more terms in the denominator that are added or subtracted is never this simple! It always involved a more complicated process that inevitably changes the numbers and coefficients in the final answer. So experienced math students can eliminate these two answer choices instantly as soon as they look at them.

Then, looking at the other two answer choices and , experienced math students see that the only difference between them is the sign in the middle, before the imaginary part. So the only step they have to do is to multiply the numerator by the complex conjugate of the denominator, . As soon as they see that the imaginary part of the numerator will have to be negative, , they already instantly know that must be wrong, and must be right. (See the **Pro Tip** above.) This is how the top SAT math students are able to answer difficult questions like this one so quickly and correctly.

### Example Question #8 : Working With Imaginary Numbers

Which of the following complex numbers is equivalent to ? (Note: )

**Possible Answers:**

**Correct answer:**

This is the classic more difficult type of complex number/imaginary number question on the SAT math section: Division of one complex number by another complex number. The most difficult and critical key step of the solution process is the very first step: You must multiply both the numerator and denominator by what we call the complex conjugate of the denominator, which is a fancy term that just means you change the sign in the middle of the complex number (before the imaginary part). The reason we do this is that this will *make the imaginary part in the denominator disappear!* Recall the difference of squares formula in algebra: . This formula works because when you FOIL , the two middle terms will cancel each other out. Well, the same thing happens here when you multiply the complex number in the denominator by its complex conjugate: the imaginary number middle terms will cancel each other out! So . The middle terms cancel each other out, leaving us with . Now you have to know that because , therefore So , and now the negative signs in the second term cancel out, making it positive: . The point of this whole process is that the end result is to simplify the denominator to a single real number, which in this case is .

To finish solving the question, now you must also multiply the original numerator by the complex conjugate of the denominator: . The two middle terms, the imaginary parts, combine: .

**Pro Tip:** If you are alert here, you can already see that the imaginary part of the final answer will have in the numerator and in the denominator! This alone eliminates all of the wrong answer choices, so you can already see now that the **correct answer choice** must be !

If you do need to finish the solution and know the numerator of the real part as well, you continue , since you must know that as we stated above. Finishing the simplification of the numerator, we get . Now we see how the entire **correct answer choice** is right.

You may wonder how experienced math students can finish solving questions so fast and complete both SAT math sections within the strict time limits and still get perfect 800 SAT Math scores. The answer is, they use the four answer choices to guide them and speed up the process dramatically. Look at the answer choices and . Do you see how they have exactly the same numbers and coefficients as the numerator and denominator of the expression in the original question? Well, experienced math students know that simplification of a fraction like this one with two or more terms in the denominator that are added or subtracted is never this simple! It always involved a more complicated process that inevitably changes the numbers and coefficients in the final answer. So experienced math students can eliminate these two answer choices instantly as soon as they look at them.

Then, looking at the other two answer choices and , experienced math students see that the only difference between them is the sign in the middle, before the imaginary part. So the only step they have to do is to multiply the numerator by the complex conjugate of the denominator, . As soon as they see that the imaginary part of the numerator will have to be negative, , they already instantly know that must be wrong, and must be right. (See the **Pro Tip** above.) This is how the top SAT math students are able to answer difficult questions like this one so quickly and correctly.

### Example Question #9 : Working With Imaginary Numbers

If the expression is rewritten in the form , where and are real numbers, what is the value of ? (Note: )

**Possible Answers:**

**Correct answer:**

The twist in this question is that you must be sure to **answer the question they are actually asking you! **The question does not ask you to provide the entire value of the complex number that you get from solving the division problem--it only asks for the** real number part** of that complex number, which is called . (Please note: Even though itself is also a real number, as the question states, in the expression this value is the coefficient of , so it is part of the imaginary number part of the expression.) Therefore, after you perform the division of complex numbers correctly, you have to focus only on

**the real number part of your answer**, not on the imaginary number part.

But first, you do have to perform the division of complex numbers correctly. The most difficult and critical key step of this process is the very first step: You must multiply both the numerator and denominator by what we call the complex conjugate of the denominator, which is a fancy term that just means you change the sign in the middle of the complex number (before the imaginary part). The reason we do this is that this will make the imaginary part in the denominator disappear! Recall the difference of squares formula in algebra: . This formula works because when you FOIL , the two middle terms will cancel each other out. Well, the same thing happens here when you multiply the complex number in the denominator by its complex conjugate: the imaginary number middle terms will cancel each other out! So . The middle terms cancel each other out, leaving us with . Now you have to know that because , therefore . So , and now the negative signs in the second term cancel out, making it positive: . The point of this whole process is that the end result is to simplify the denominator to a single real number, which in this case is .

To continue the complex number division process, now you must also multiply the original numerator by the complex conjugate of the denominator: . The two middle terms, the imaginary parts, combine: . Continuing, you must know that as we stated above, so . Finishing the simplification of the numerator, we get .

Conveniently, we now see that our numerator and our denominator (see above) simplify dramatically! . Wow! (Note: On the SAT, these things never happen by accident. The test question writers carefully design the questions so that they often simplify dramatically like this, when the student solves the questions correctly.)

Now, by solving the complex number division process correctly, we have rewritten the original expression in the form , which in this case has the value . Recall that you must **answer the question they are actually asking you**, and the question asks you for the value of . Therefore the **correct answer choice** is .

**Pro Tip:** Experienced math students may be able to save a little time during the solution process: If you know that the question is only asking for the real number part of the final complex number, then when you solve the numerator above, you can skip the imaginary number parts, which are the two middle terms when you FOIL . So you can just solve the terms that will become the real number parts: . Since this is the real number part of the numerator, and the whole denominator is also , now you already know the final real number part will be , which is what the question is actually asking for, so that is the final correct answer choice for this question.

### Example Question #10 : Working With Imaginary Numbers

If the expression is rewritten in the form , where and are real numbers, what is the value of ? (Note: )

**Possible Answers:**

**Correct answer:**

The twist in this question is that you must be sure to **answer the question they are actually asking you**! The question does not ask you to provide the entire value of the complex number that you get from solving the division problem--it only asks for the **coefficient of the imaginary number part** of that complex number, which is called . (Please note: Even though itself is a real number, as the question states, in the expression this value b is the coefficient of , so it is part of the imaginary number part of the expression.) Therefore, after you perform the division of complex numbers correctly, you have to focus only on **the coefficient of the imaginary number part of your answer**, not on the real number part.

But first you do have to perform the division of complex numbers correctly. The most difficult and critical key step of this process is the very first step: You must multiply both the numerator and denominator by what we call the complex conjugate of the denominator, which is a fancy term that just means you change the sign in the middle of the complex number (before the imaginary part). The reason we do this is that this will make the imaginary part in the denominator disappear! Recall the difference of squares formula in algebra: . This formula works because when you FOIL , the two middle terms will cancel each other out. Well, the same thing happens here when you multiply the complex number in the denominator by its complex conjugate: the imaginary number middle terms will cancel each other out! So . The middle terms cancel each other out, leaving us with . Now you have to know that because , therefore . So , and now the negative signs in the second term cancel out, making it positive: . The point of this whole process is that the end result is to simplify the denominator to a single real number, which in this case is . To continue the complex number division process, now you must also multiply the original numerator by the complex conjugate of the denominator: . The two middle terms, the imaginary parts, combine: . Continuing, you must know that as we stated above, so . Finishing the simplification of the numerator, we get .

Conveniently, we now see that our numerator and our denominator (see above) simplify dramatically! . Wow! (Note: On the SAT, these things never happen by accident. The test question writers carefully design the questions so that they often simplify dramatically like this, when the student solves the questions correctly.) Now, by solving the complex number division process correctly, we have rewritten the original expression in the form which in this case has the value . Recall that you must **answer the question they are actually asking you**, and the question asks you for the value of . Therefore the **correct answer choice** is . (Note: “” is the same value as “”, so the coefficient is , even though the number is usually not written before the “” in this case since it is not necessary.)

Pro Tip: Experienced math students may be able to save a little time during the solution process: If you know that the question is only asking for the coefficient of the imaginary number part of the final complex number, then when you solve the numerator above, you can skip the real number parts, which are the first and last terms when you FOIL . So you can just solve the terms that will become the imaginary number parts: . Since this is the imaginary number part of the numerator, and the whole denominator is , now you already know the final imaginary number part will be , and its coefficient is , which is what the question is actually asking for, so that is the final correct answer choice for this question.

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