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SAT Math : How to add complex numbers

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Example Questions

Example Question #11 : Squaring / Square Roots / Radicals

Simplify: $\sqrt{-3}+\sqrt{-9}+\sqrt{-16}$

Possible Answers:

$2i\sqrt7$

$i\sqrt3 +7i$

$i\sqrt3-i$

$i\sqrt3 +i$

$i\sqrt3 -7i$

Correct answer:

$i\sqrt3 +7i$

Explanation:

Rewrite $\sqrt{-3}+\sqrt{-9}+\sqrt{-16}$ in their imaginary terms.

$i=\sqrt{-1}$

$\sqrt{-3}+\sqrt{-9}+\sqrt{-16}=i\sqrt3+3i+4i = i\sqrt3 +7i$

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Example Question #1 : Complex Numbers

Add $5 + i \sqrt {5 }$ and its complex conjugate.

Possible Answers:

$2 i \sqrt{5}$

$-2 i \sqrt{5}$

Correct answer:

Explanation:

The complex conjugate of a complex number a+ bi is a - bi . Therefore, the complex conjugate of $5 + i \sqrt {5 }$ is $5 - i \sqrt {5 }$ ; add them by adding real parts and adding imaginary parts, as follows:

$(5 + i \sqrt {5 } )+( 5 - i \sqrt {5 })$

$= 5 + 5 + i \sqrt {5 } - i \sqrt {5 }$

= 10 ,

the correct response.

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Example Question #3 : Complex Numbers

Add $13 - i \sqrt{13}$ to its complex conjugate.

Possible Answers:

$2i \sqrt{13}$

$-2i \sqrt{13}$

Correct answer:

Explanation:

The complex conjugate of a complex number a - bi is a+ bi . Therefore, the complex conjugate of $13 - i \sqrt{13}$ is $13 + i \sqrt{13}$ ; add them by adding real parts and adding imaginary parts, as follows:

$(13 - i \sqrt{13} )+( 13 + i \sqrt{13})$

$= 13+13 - i \sqrt {13 }+ i \sqrt {13}$

= 26

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Example Question #2 : Complex Numbers

An arithmetic sequence begins as follows:

3, 3i, ...

Give the next term of the sequence

Possible Answers:

3+ 6i

3- 6i

-3 + 6i

Correct answer:

-3 + 6i

Explanation:

The common difference of an arithmetic sequence can be found by subtracting the first term from the second:

$d = a_{2} - a_{1}$

d = 3i - 3 = -3 + 3i

Add this to the second term to obtain the desired third term:

$a_{3} = a_{2} +d = 3i + ( -3 + 3i) = -3 + 3i + 3i = -3 + 6i$ .

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Example Question #1 : Complex Numbers

Simplify: (3+5i)+(4-2i)+(-2+i)

Possible Answers:

5+4i

5-4i

9+8i

9-8i

10-9i

Correct answer:

5+4i

Explanation:

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):

(3+5i)+(4-2i)+(-2+i)

=3+4+(-2)+5i+(-2i)+i

=5+4i

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Example Question #2 : Complex Numbers

For $i=\sqrt{-1}$ , what is the sum of $\frac{3}{2} + \frac{1}{4} i$ and its complex conjugate?

Possible Answers:

$\frac{1}{2} i$

$-\frac{1}{2} i$

$\frac{7}{4} + \frac{7}{4} i$

Correct answer:

Explanation:

The complex conjugate of a complex number a+bi is a - bi , so $\frac{3}{2} + \frac{1}{4} i$ has $\frac{3}{2} - \frac{1}{4} i$ as its complex conjugate. The sum of the two numbers is

$\left (\frac{3}{2} + \frac{1}{4} i \right ) + \left ( \frac{3}{2} - \frac{1}{4} i \right )$

$= \frac{3}{2} + \frac{1}{4} i + \frac{3}{2} - \frac{1}{4} i$

$= \frac{3}{2} + \frac{3}{2} + \frac{1}{4} i - \frac{1}{4} i$

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Example Question #3 : How To Add Complex Numbers

Evaluate:

$9i^{9}+ 10 i^{10}+ 11 i^{11}+ 12 i^{12}$

Possible Answers:

2 +2i

-2 - 2i

None of these

2 - 2i

-2+ 2i

Correct answer:

2 - 2i

Explanation:

A power of can be evaluated by dividing the exponent by 4 and noting the remainder. The power is determined according to the following table:

$\begin{matrix} \textrm{\underline{Rem}}& \textrm{\underline{Power}} \\ 0& 1\\ 1& i\\ 2& -1\\ 3& -i \end{matrix}$

$9 \div 4 = 2 \textrm{ R }1$ , so $i^{9} = i$

$10 \div 4 = 2 \textrm{ R }2$ , so $i^{10} = -1$

$11 \div 4 = 2 \textrm{ R }3$ , so $i^{11} = - i$

$12 \div 4 = 3 \textrm{ R }0$ , so $i^{12} = 1$

Substituting:

$9i^{9}+ 10 i^{10}+ 11 i^{11}+ 12 i^{12}$

= 9(i) + 10 (-1)+ 11 (-i)+ 12 (1)

= 9i - 10 - 11i+ 12

Collect real and imaginary terms:

- 10 + 12 + 9i - 11i

= 2 - 2i

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Example Question #592 : Algebra

Evaluate:

$i^{6} + i^{7}+ i^{8}+ i^{9}$

Possible Answers:

-2i

Correct answer:

Explanation:

A power of can be evaluated by dividing the exponent by 4 and noting the remainder. The power is determined according to the following table:

$\begin{matrix} \textrm{\underline{Rem}}& \textrm{\underline{Power}} \\ 0& 1\\ 1& i\\ 2& -1\\ 3& -i \end{matrix}$

$6 \div 4 = 1 \textrm{ R }2$ , so $i^{6} = -1$

$7 \div 4 = 1 \textrm{ R }3$ , so $i^{7} = -i$

$8 \div 4 = 2 \textrm{ R }0$ , so $i^{8} = 1$

$9 \div 4 = 2 \textrm{ R }1$ , so $i^{9} = i$

Substituting:

$i^{6} + i^{7}+ i^{8}+ i^{9}$

= -1+ ( - i )+ 1 +i

= -1 + 1 - i + i

= 0

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SAT Math : How to add complex numbers

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #11 : Squaring / Square Roots / Radicals

Example Question #1 : Complex Numbers

Example Question #3 : Complex Numbers

Example Question #2 : Complex Numbers

Example Question #1 : Complex Numbers

Example Question #2 : Complex Numbers

Example Question #3 : How To Add Complex Numbers

Example Question #592 : Algebra

All SAT Math Resources

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