How to add complex numbers - SAT Math

Card 0 of 8

Question

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

Answer

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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Question

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

Answer

The complex conjugate of a complex number is , 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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Question

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

Answer

The complex conjugate of a complex number is . 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:

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

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

,

the correct response.

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Question

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

Answer

The complex conjugate of a complex number is . 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:

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

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

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

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Question

An arithmetic sequence begins as follows:

Give the next term of the sequence

Answer

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

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

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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Question

Evaluate:

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

Answer

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:

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

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Question

Evaluate:

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

Answer

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:

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

Collect real and imaginary terms:

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Question

Simplify:

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):

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