How to graph complex numbers - SSAT Upper Level Quantitative

Card 0 of 92

Define an operation $\wedge$ as follows:

For all complex numbers ,

$a \wedge b = a \div (b+ 2i)$

Evaluate $8 \wedge 4$

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$a \wedge b = a \div (b+ 2i) = $\frac{a}{b+2i}$$

$8 \wedge 4 = $\frac{8}{4+2i}$$

Multiply both numerator and denominator by the conjugate of the denominator, , to rationalize the denominator:

$8 \wedge 4 = $\frac{8(4-2i)}{(4+2i)(4-2i)}$$

$= $\frac{8\cdot 4-8\cdot $2i}{4^{2}$$$+2^{2}$}$

$= $\frac{32-16i}{16+4}$$

$= $\frac{32-16i}{20}$$

$= $\frac{32 }{20}$ - $\frac{ 16}{20}$i$

$= $\frac{8}{5}$ -$\frac{ 4}{5}$i$

Raise to the power of 4.

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Multiply the following complex numbers:

$\left (5 + i \right )\left (5 + 2i \right )$

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FOIL the product out:

$\left (5 + i \right )\left (5 + 2i \right )$

To FOIL multiply the first terms from each binomial together, multiply the outer terms of both terms together, multiply the inner terms from both binomials together, and finally multiply the last terms from each binomial together.

$=5 \cdot 5 + 5 \cdot 2i + 5 \cdot i + i\cdot 2i$

$=25 + 10i + 5 i + $2i^{2}$$

Recall that i is an imaginary number and by definition . Substituting this into the function is as follows.

Multiply:

$\left (7 - 2i \right )\left (7 - 3i \right )$

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FOIL the product out:

$\left (7 - 2i \right )\left (7 - 3i \right )$

$= 7 \cdot 7 - 7 \cdot 3i - 7 \cdot 2i + 2i \cdot 3i$

$= 49 -21i - 14i + 6 $i^{2}$$

Simplify:

$\left ( 8 - 3i\right ) ^{2}$

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Use the square of a binomial pattern to multiply this:

$\left ( 8 - 3i\right ) ^{2}$

$= 8 ^{2}- 2 \cdot8\cdot3i + \left ( 3i\right $)^{2}$$

$= 64- 48i + $3^{2}$$i^{2}$$

Multiply:

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This is a product of an imaginary number and its complex conjugate, so it can be evaluated using this formula:

$\left (a + bi \right )\left (a -bi \right ) = $a^{2}$ + b ^{2}$

$\left (1.1 + 0.6i \right )\left (1.1 -0.6 i \right ) = $1.1^{2}$ + 0.6 ^{2} = 1.21 + 0.36 = 1.57$

Multiply:

$\left (5 - 3i \right )\left (5 +3i \right )$

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This is a product of an imaginary number and its complex conjugate, so it can be evaluated using this formula:

$\left (a - bi \right )\left (a +bi \right ) = $a^{2}$ + b ^{2}$

$\left (5 - 3i \right )\left (5 +3i \right ) = 5 ^{2} + 3 ^{2}= 25 + 9 = 34$

Evaluate .

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$= $\frac{1}{243 i }$ \cdot = $\frac{1\cdot i}{243 i \cdot i }$ = $\frac{i}{243 (-1)}$ = - $\frac{1}{243}$ i$

Define an operation $\wedge$ as follows:

For all complex numbers ,

$a \wedge b = a \div (b+ 3)$

Evaluate $7i \wedge 5i$

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$a \wedge b = a \div (b+ 3) = $\frac{a}{b+3}$$

$7i \wedge 5i = $\frac{7i }{5i+3}$ = $\frac{7i }{3+ 5i}$$

Multiply both numerator and denominator by the conjugate of the denominator, , to rationalize the denominator:

$7i \wedge 5i = $\frac{7i (3-5i)}{(3+ 5i)(3-5i)}$$

$= $\frac{7i \cdot 3-7i \cdot $5i}{3^{2}$$+ $5^{2}$}$

$= $\frac{21i -35 \cdot $i^{2}$$}{9+25}$

$= $\frac{21i-35 (-1)}{34}$$

$= $\frac{21i+35}{34}$$

$= $\frac{ 35+21i}{34}$$

$= $\frac{ 35 }{34}$ + $\frac{ 21}{34}$ i$

Define an operation as follows:

For all complex numbers ,

$ab = a ^${5}-b^{3}$$

Evaluate $\left (2 i \right ) \left (3i \right )$

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$ab = a ^${5}-b^{3}$$

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

$= $2^{5}$ $i^{5}$ - $3^{3}$ i ^{3}$

$= 32\cdot $i^{4}$ \cdot i - 27 \cdot (-i )$

$= 32\cdot 1 \cdot i +27 i$

Define an operation as follows:

For all complex numbers ,

Evaluate

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$= 4+7i + 3i \cdot 3- 3i \cdot 8i$

$= 4+7i + 9i - 24i ^{2}$

Give the number which, when multiplied by , yields the same result as if it were increased by 6.

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Let be the number in question. The statement "\[a number\] multiplied by yields the same result as if it were increased by 6" can be written as

We can solve this for as follows:

$N(-1+i) \div (-1+i) = 6 \div (-1+i)$

$N = \frac{6}{-1+i}$

Rationalize the denominator by multiplying both numerator and denominator by the conjugate of the denominator, which is :

$N = \frac{6(-1- i)}{(-1+i)(-1- i)}$

$= \frac{6(-1)- 6\cdot i}{(-1)^{2}+1^{2}}$

$= \frac{-6-6i}{1+1}$

$= \frac{-6-6i}{2}$

Give the number which, when added to 20, yields the same result as if it were subtracted from .

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Let be the number in question. The statement "\[a number\] added to 20 yields the same result as if it were aubtracted from " can be written as

Solve for :

$\frac{1}{2} \cdot 2N = \frac{1}{2} \cdot (-20 + 14 i)$

Define an operation $\bigstar$ as follows:

For all complex numbers ,

$\bigstar a = $a^{2}$ + a$

Evaluate $\bigstar (6i)$ .

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$\bigstar a = $a^{2}$ + a$

$\bigstar (6i) = $(6i)^{2}$ + 6i$

$= $6^{2}$ $i^{2}$ + 6i$

Define an operation $\bigstar$ as follows:

For all complex numbers ,

$\bigstar a = $a^{2}$i + a$

Evaluate $\bigstar (7i)$ .

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$\bigstar a = $a^{2}$i + a$

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

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

$= 49\cdot (-1) \cdot i + 7i$

Define an operation $\wedge$ as follows:

For all complex numbers ,

$a \wedge b = a \div (b+ 4)$ .

If $(8i )\wedge N = 16$ , evaluate .

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$(8i )\wedge N = 16$ ,

by our definition, can be rewritten as

$8i \div (N+ 4) = 16$

or

$$\frac{8i }{N+ 4}$= 16$

Taking the reciprocal of both sides, then multiplying:

$\frac{N+ 4}{8i }$= $\frac{1}{16}$

$$\frac{N+ 4}{8i }$ \cdot 8i = $\frac{1}{16}$\cdot 8i$

$N+ 4 = $\frac{1}{2}$i$

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

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

Define an operation $\bigstar$ as follows:

For all complex numbers ,

$\bigstar a = $a^{2}$i - 4 a$

If $\bigstar N = 0$ and $N \ne 0$ , evaluate .

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If $\bigstar N = 0$ , then

.

Distribute out to yield

Either or . However, we are given that $N \ne 0$ , so

$\frac{N i }{i}$= $\frac{4}{i}$

$N= $\frac{4}{i}$ = $\frac{4 \cdot i}{i\cdot i}$= $\frac{4i}{-1}$ = -4i$

Multiply the complex conjugate of by . What is the result?

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The complex conjugate of a complex number is . Since , its complex conjugate is .

Multiply this by :

Recall that by definition .

$-7i \cdot 3i = -7 \cdot 3 \cdot i \cdot i = -21 (-1) = 21$

Multiply the complex conjugate of 8 by . What is the result?

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The complex conjugate of a complex number is . Since , its complex conjugate is itself. Multiply this by :

$8 \cdot 5i = 40 i$

Multiply the complex conjugate of by . What is the result?

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The complex conjugate of a complex number is , so the complex conjugate of is . Multiply this by :

$(3-7i) \cdot 4i = 3 \cdot 4i - 7 i \cdot 4i$

$= 12i - 7 \cdot 4 \cdot i \cdot i$

$= 12i -28 \cdot (-1)$