How to multiply complex numbers - SAT Math

Card 0 of 192

Question

$x = 5+ i \sqrt{7}$ ; is the complex conjugate of .

Evaluate

$x^{2} - 2xy + y^{2}$ .

Answer

$x^{2} - 2xy + y^{2}$ conforms to the perfect square trinomial pattern

$x^{2} - 2xy + y^{2} = (x-y) ^{2}$ .

The easiest way to solve this problem is to subtract and , then square the difference.

The complex conjugate of a complex number is .

$x = 5+ i \sqrt{7}$ ,

so is the complex conjugate of this;

$x = 5- i \sqrt{7}$

$x-y = (5+ i \sqrt{7}) - (5- i \sqrt{7}) = 5- 5 +i \sqrt{7}+ i \sqrt{7} = 2i \sqrt{7}$

$x^{2} - 2xy + y^{2} = (x-y) ^{2}$

$=( 2i \sqrt{7} )^{2}$

Taking advantage of the Power of a Product Rule and the fact that $i^{2} = -1$ :

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

$=4 \cdot (-1 )\cdot 7$

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Question

Find the product of (3 + 4i)(4 - 3i) given that i is the square root of negative one.

Answer

Distribute (3 + 4i)(4 - 3i)

3(4) + 3(-3i) + 4i(4) + 4i(-3i)

12 - 9i + 16i -12i2

12 + 7i - 12(-1)

12 + 7i + 12

24 + 7i

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Question

$x^{4}-1$ has 4 roots, including the complex numbers. Take the product of $\left ( 5+2i \right )$ with each of these roots. Take the sum of these 4 results. Which of the following is equal to this sum?

Answer

$x^{4}-1=\left ( x^{2} -1\right )\left ( x^{2}+1 \right )=\left ( x+1 \right )\left ( x-1 \right )\left ( x-i \right )\left ( x+i \right )$

This gives us roots of

$1,\ -1,\ i,\ -i$

The product of $\left ( 5+2i \right )$ with each of these gives us:

$(5+2i)\cdot i = 5i + 2i^2 = 5i +2(-1) = 5i-2 = -2 + 5i$

$(5+2i)\cdot (-i) = -5i -2i^2 = -5i -2(-1) = -5i +2 = 2-5i$

$(5+2i)\cdot 1 = 5+2i$

$(5+2i)\cdot (-1) = -5 -2i$

The sum of these 4 is:

What we notice is that each of the roots has a negative. It thus makes sense that they will all cancel out. Rather than going through all the multiplication, we can instead look at the very beginning setup, which we can simplify using the distributive property:

$1(5+2i) + (-1)\cdot(5+2i)+ i(5+2i)+(-i) \cdot(5+2i) = (5+2i)\cdot[1+(-1)+i+(-i)]$

$= (5+2i)\cdot[0] = 0$

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Question

Simplify:

$(3i)^{6}$

Answer

Apply the Power of a Product Property:

$(3i)^{6} = 3 ^{6} \cdot i^{6} = 729 \cdot i^{6}$

A power of can be found by dividing the exponent by 4 and noting the remainder. 6 divided by 4 is equal to 1, with remainder 2, so

$i^{6} = i^{2} = -1$

Substituting,

$(3i)^{6} = 729 \cdot (-1) = -729$ .

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Question

Multiply $3 + i \sqrt {3 }$ by its complex conjugate.

Answer

The complex conjugate of a complex number is . The product of the two is the number

$a^{2}+b^{2}$ .

Therefore, the product of $3 + i \sqrt {3 }$ and its complex conjugate $3 - i \sqrt {3 }$ can be found by setting and $b = \sqrt{3}$ in this pattern:

$a^{2}+b^{2} = 3^{2}+ (\sqrt{3})^{2} = 9 + 3 = 12$ ,

the correct response.

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Question

Multiply $11 - i \sqrt{11}$ by its complex conjugate.

Answer

The complex conjugate of a complex number is . The product of the two is the number

$a^{2}+b^{2}$ .

Therefore, the product of $11 - i \sqrt{11}$ and its complex conjugate $11 + i \sqrt{11}$ can be found by setting and $b = \sqrt{11}$ in this pattern:

$a^{2}+b^{2} = 11^{2}+ (\sqrt{11})^{2} = 121+11 = 132$ ,

the correct response.

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Question

Evaluate:

$i^{-31}$

Answer

$a^{-m}$ 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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Question

What is the product 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 product of and is equal to $a^{2}+ b^{2}$ , so set $a = \frac{3}{2} , b= \frac{1}{4}$ in this expression, and evaluate:

$a^{2}+ b^{2} = \left ( \frac{3}{2} \right )^{2} + \left (\frac{1}{4} \right )^{2} = \frac{9}{4} + \frac{1}{16} = \frac{37}{16}$ .

This is not among the given responses.

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Question

Multiply and simplify:

$\sqrt{-35} \cdot \sqrt{-5}$

Answer

The two factors are both square roots of negative numbers, and are therefore imaginary. Write both in terms of before multiplying:

$\sqrt{-35} = i \sqrt{35}$

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

Therefore, using the Product of Radicals rule:

$\sqrt{-35} \cdot \sqrt{-5}$

$= i \sqrt{35} \cdot i \sqrt{5}$

$= i \cdot i \cdot \sqrt{35} \cdot \sqrt{5}$

$= i ^{2}\cdot \sqrt{35 \cdot 5 }$

$=-1 \cdot \sqrt{175 }$

$=-1 \cdot \sqrt{25 }\cdot \sqrt{7}$

$=-1 \cdot 5\cdot \sqrt{7}$

$=-5\sqrt{7}$

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Question

Evaluate $x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}$

Answer

$x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}$ is recognizable as the cube of the binomial . That is,

$x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}= (x-y) ^{3}$

Therefore, setting and and evaluating:

$x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}=[ (7+4i) - (7-4 i) ]^{3}$

$= (7-7 +4i+4i ) ^{3}$

$=(8i)^{3}$

Applying the Power of a Product Rule and the fact that $i^{3} = -i$ :

$(8i)^{3} = 8 ^{3} i ^{3} = 512 (-i) = -512i$ ,

the correct value.

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Question

Evaluate $x^{3} + 3x^{2} y + 3 xy^{2} + y^{3}$

Answer

$x^{3} + 3x^{2} y + 3 xy^{2} + y^{3}$ is recognizable as the cube of the binomial . That is,

$x^{3} + 3x^{2} y + 3 xy^{2} + y^{3} = (x+y) ^{3}$

Therefore, setting and and evaluating:

$x^{3} + 3x^{2} y + 3 xy^{2} + y^{3} =[ (6+3i) + (6-3i) ]^{3}$

$= (6+6+3i -3i) ^{3}$

$=12 ^{3}$

.

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Question

Raise to the third power.

Answer

To raise any expression to the third power, use the pattern

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

Setting :

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

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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Question

Raise to the fourth power.

Answer

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 by squaring , then squaring the result.

Using the binomial square pattern to square :

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

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:

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

Square this using the same steps:

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

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

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

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Question

; is the complex conjugate of .

Evaluate

$x^{2} + 2xy + y^{2}$ .

Answer

$x^{2} + 2xy + y^{2}$ conforms to the perfect square trinomial pattern

$x^{2} + 2xy + y^{2} = (x+y) ^{2}$ .

The easiest way to solve this problem is to add and , then square the sum.

The complex conjugate of a complex number is .

,

so is the complex conjugate of this;

,

and

Substitute 8 for :

$x^{2} + 2xy + y^{2} = (x+y) ^{2} = 8^{2} = 64$ .

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Question

; is the complex conjugate of .

Evaluate

$x^{2} - 2xy + y^{2}$ .

Answer

$x^{2} - 2xy + y^{2}$ conforms to the perfect square trinomial pattern

$x^{2} - 2xy + y^{2} = (x-y) ^{2}$ .

The easiest way to solve this problem is to subtract and , then square the difference.

The complex conjugate of a complex number is .

,

so is the complex conjugate of this;

Substitute for :

$x^{2} - 2xy + y^{2} = (x-y) ^{2} =( 16 i)^{2} = 16 ^{2} \cdot i^{2}$

By definition, $i^{2} = -1$ , so, substituting,

$16 ^{2} \cdot i^{2} = 256 (-1 ) = -256$ ,

the correct choice.

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Question

Raise $8 + i \sqrt{5}$ to the power of 4.

Answer

The easiest way to find $\left (8 + i \sqrt{5} \right )^{4}$ is to note that

$\left (8 + i \sqrt{5} \right )^{4} =\left [ \left (8 + i \sqrt{5} \right )^{2} \right ]^{2}$ .

Therefore, we can find the fourth power of $8 + i \sqrt{5}$ by squaring $8 + i \sqrt{5}$ , then squaring the result.

Using the binomial square pattern to square $8 + i \sqrt{5}$ :

$(8 + i \sqrt{5} )^{2} = 8^{2} + 2 \cdot 8 \cdot i \sqrt{5} + (i \sqrt{5} ) ^{2}$

Applying the Power of a Product Property:

$(8 + i \sqrt{5} )^{2} = 8^{2} + 2 \cdot 8 \cdot i \sqrt{5} + i^{2} \cdot \left ( \sqrt{5} \right) ^{2}$

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

$(8 + i \sqrt{5} )^{2} = 8^{2} + 2 \cdot 8 \cdot i \sqrt{5} + (-1) \cdot 5$

$(8 + i \sqrt{5} )^{2} = 64 +16 i \sqrt{5} -5$

$= 59 +16 i \sqrt{5}$

Square this using the same steps:

$(59 +16 i \sqrt{5} )^{2} = 59^{2} + 2 \cdot 59 \cdot 16 i \sqrt{5} + (16i \sqrt{5} ) ^{2}$

$= 59^{2} + 2 \cdot 59 \cdot 16 i \sqrt{5} + 16^{2} \cdot i^{2} \cdot (\sqrt{5} ) ^{2}$

$=3,481+1,888 i \sqrt{5} + 256 \cdot (-1) \cdot 5$

$=3,481+1,888 i \sqrt{5} -1,280$

$=2,201+1,888 i \sqrt{5}$ ,

the correct response.

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Question

Raise $6- i\sqrt{7}$ to the power of 3.

Answer

To raise any expression to the third power, use the pattern

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

Setting $A = 6, B = i \sqrt{7}$ :

$(6- i \sqrt{7} )^{3} = 6^{3} - 3\cdot 6^{2} \cdot i \sqrt{7} + 3 \cdot 6 \cdot ( i \sqrt{7})^{2} - ( i \sqrt{7})^{3}$

Taking advantage of the Power of a Product Rule:

$(6- i \sqrt{7} )^{3} = 6^{3} - 3\cdot 6^{2} \cdot i \sqrt{7} + 3 \cdot 6 \cdot i^{2} \cdot ( \sqrt{7})^{2} - i ^{3}\cdot (\sqrt{7})^{3}$

Since $i^{2} = -1$ ,

and

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

$(6- i \sqrt{7} )^{3}=216 - 3\cdot 36 \cdot i \sqrt{7} + 3 \cdot 6 \cdot 7 \cdot (-1) - (-i) \cdot 7\cdot \sqrt{7}$

$(6- i \sqrt{7} )^{3} =216 - 108i \sqrt{7} -126 + 7i\sqrt{7}$

Collecting real and imaginary terms:

$(6- i \sqrt{7} )^{3} =216 -126 - 108i \sqrt{7} + 7i\sqrt{7}$

$(6- i \sqrt{7} )^{3} =90- 101i \sqrt{7}$

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Question

Evaluate $\left ( 4i\right )^{3} \cdot (3i)^{4}$ .

Answer

Apply the Power of a Product Rule:

$\left ( 4i\right )^{3} \cdot (3i)^{4}$

$= 4^{3}\cdot i^{3} \cdot 3^{4} \cdot i^{4}$

$i^{3} = i^{2} \cdot i = (-1 ) \cdot i = -i$ ,

and

$i^{4} = (i^{2})^{2} = (-1)^{2} = 1$ ,

so, substituting and evaluating:

$4^{3}\cdot i^{3} \cdot 3^{4} \cdot i^{4}$

$= 64 \cdot (-i) \cdot 81 \cdot 1$

$= - 64 \cdot 81 \cdot 1 \cdot i$

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Question

Evaluate $\left ( 2 i \sqrt{5}\right )^{5} \cdot \left ( 3 i \sqrt{5} \right )^{3}$

Answer

Apply the Power of a Product Rule:

$\left ( 2 i \sqrt{5}\right )^{5} \cdot \left ( 3 i \sqrt{5} \right )^{3}$

$= 2 ^{5} \cdot i^{5} \cdot \left ( \sqrt{5}\right )^{5} \cdot 3 ^{3} \cdot i ^{3} \cdot \left ( \sqrt{5} \right )^{3}$

$= 2 ^{5} \cdot 3 ^{3} \cdot i^{5}\cdot i ^{3} \cdot \left ( \sqrt{5}\right )^{5} \cdot \left ( \sqrt{5} \right )^{3}$

Applying the Product of Powers Rule:

$2 ^{5} \cdot 3 ^{3} \cdot i^{5}\cdot i ^{3} \cdot \left ( \sqrt{5}\right )^{5} \cdot \left ( \sqrt{5} \right )^{3}$

$= 2 ^{5} \cdot 3 ^{3} \cdot i^{5+3} \cdot \left ( \sqrt{5}\right )^{5+3}$

$= 2 ^{5} \cdot 3 ^{3}\cdot \left ( \sqrt{5}\right )^{8} \cdot i^{8}$

raised to any multiple of 4 is equal to 1, and $\sqrt{5} = 5 ^{\frac{1}{2}}$ , so, substituting and evaluating:

$2 ^{5} \cdot 3 ^{3}\cdot \left ( \sqrt{5}\right )^{8} \cdot i^{8}$

$= 2 ^{5} \cdot 3 ^{3}\cdot \left ( 5^{\frac{1}{2}} \right )^{8} \cdot 1$

$= 2 ^{5} \cdot 3 ^{3}\cdot 5^{ \frac{1}{2} \cdot 8} \cdot 1$

$= 2 ^{5} \cdot 3 ^{3}\cdot 5^{4} \cdot 1$

$= 32 \cdot 27 \cdot 625 \cdot 1$

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Question

Raise to the fourth power.

Answer

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 to the fourth power is to square it, then to square the result.

Using the binomial square pattern to square :

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

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:

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

Square this using the same steps:

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

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

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

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