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SAT Math : How to multiply complex numbers

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Example Questions

Example Question #32 : Complex Numbers

Raise 7+2i to the fourth power.

Possible Answers:

1,241 +2,520 i

None of these

2,809 +2,520 i

2,025+2,968i

3,593+2,968i

Correct answer:

1,241 +2,520 i

Explanation:

By the Power of a Power Rule, the fourth power of any number is equal to the square of the square of that number:

$x^{4} = x ^{2 \cdot 2} =( x^{2})^{2}$

Therefore, one way to raise 7+2i to the fourth power is to square it, then to square the result.

Using the binomial square pattern to square 7+2i :

$(7+2i)^{2} = 7^{2} + 2 \cdot 7 \cdot 2i + (2i ) ^{2}$

Applying the Power of a Product Property:

$(7+2i)^{2} = 7^{2} + 2 \cdot 7 \cdot 2i + 2^{2} \cdot i ^{2}$

Since $i^{2} = -1$ by definition:

$(7+2i)^{2} = 7^{2} + 2 \cdot 7 \cdot 2i + 2^{2} \cdot (-1)$

= 49 + 28i -4

= 45 + 28i

Square this using the same steps:

$(45 + 28i )^{2} = 45 ^{2} + 2 \cdot 45 \cdot 28 i + (28 i) ^{2}$

$= 45 ^{2} + 2 \cdot 45 \cdot 28 i + 28^{2} i ^{2}$

$= 45 ^{2} + 2 \cdot 45 \cdot 28 i + 28^{2} (-1)$

= 2,025 +2,520 i -784

= 1,241 +2,520 i

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Example Question #51 : Squaring / Square Roots / Radicals

Raise 8 - 3i to the fourth power.

Possible Answers:

7,633 - 7,008i

None of these

721-5,280 i

3,025 - 7,008i

5,329-5,280 i

Correct answer:

721-5,280 i

Explanation:

The easiest way to find $(8 - 3i)^{4}$ is to note that

$(8 - 3i)^{4} =[ (8 - 3i)^{2} ]^{2}$ .

Therefore, we can find the fourth power of 8 - 3i by squaring 8 - 3i , then squaring the result.

Using the binomial square pattern to square 8 - 3i :

$(8 - 3i)^{2} = 8^{2} - 2 \cdot 8 \cdot 3i + (3i)^{2}$

Applying the Power of a Product Property:

$(8 - 3i)^{2} = 8^{2} - 2 \cdot 8 \cdot 3i + 3^{2} \cdot i^{2}$

Since $i^{2} = -1$ by definition:

$(8 - 3i)^{2} = 64 - 48 i + 9 (-1)$

= 64 - 48 i -9

= 55 - 48 i

Square this using the same steps:

$( 55 - 48 i )^{2} = 55^{2} - 2 \cdot 55 \cdot 48 i + (48i)^{2}$

$= 3,025-5,280 i + 48^{2} i^{2}$

= 3,025-5,280 i + 2,304(-1)

= 3,025-5,280 i - 2,304

= 721-5,280 i

Therefore, $(8 - 3i)^{4} = 721-5,280 i$

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Example Question #42 : Complex Numbers

Raise 5+4i to the third power.

Possible Answers:

-115 + 364i

365 + 364i

None of these

-115 + 236i

365 + 236i

Correct answer:

-115 + 236i

Explanation:

To raise any expression A+B to the third power, use the pattern

$\left (A+B \right )^{3} = A^{3} + 3A^{2} B + 3 AB^{2} + B^{3}$

Setting A = 5, B = 4i :

$(5+4i)^{3} = 5^{3} + 3\cdot 5^{2} \cdot 4i + 3 \cdot 5 \cdot (4i)^{2} + (4i)^{3}$

Taking advantage of the Power of a Product Rule:

$(5+4i)^{3} = 5^{3} + 3\cdot 5^{2} \cdot 4i + 3 \cdot 5 \cdot 4^{2} \cdot i^{2} +4 ^{3} \cdot i ^{3}$

Since $i^{2} = -1$ and $i^{3} = i^{2} \cdot i = -1 \cdot i = -i$ :

$(5+4i)^{3} = 125 + 3\cdot 25 \cdot 4 i + 3 \cdot 5 \cdot 16 \cdot (-1) +64 \cdot (- i )$

$(5+4i)^{3} = 125+ 300i -240 -64i$

Collecting real and imaginary terms:

$(5+4i)^{3} = 125 -240 + 300i -64i$

$(5+4i)^{3} = -115 + 236i$

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Example Question #21 : How To Multiply Complex Numbers

Evaluate:

$i^{-31}$

Possible Answers:

The expression is undefined

Correct answer:

Explanation:

$a^{-m}$ is defined to be equal to $\frac{1}{a^{m}}$ for any real or imaginary and for any real ; therefore,

$i^{-31} = \frac{1}{i^{31}}$

To evaluate a positive power of , divide the power by 4 and note the remainder:

$31 \div 4= 7 \textrm{ R }3$

Therefore,

$i ^{31} = i ^{3} = -i$

Substituting,

$i^{-31} = \frac{1}{-i}$

Rationalizing the denominator by multiplying both numerator and denominator by :

$i^{-31} = \frac{1 \cdot i}{-i \cdot i }= \frac{ i}{-i ^{2}}= \frac{ i}{-(-1)}= \frac{ i}{1}= i$

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SAT Math : How to multiply complex numbers

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #32 : Complex Numbers

Example Question #51 : Squaring / Square Roots / Radicals

Example Question #42 : Complex Numbers

Example Question #21 : How To Multiply Complex Numbers

All SAT Math Resources

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