Complex Numbers

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SAT Math › Complex Numbers

Questions 1 - 10

$x^{4}-1$ 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?

The correct answer is not listed.

Explanation

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

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:

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

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

Simplify:

Explanation

Use FOIL:

Combine like terms:

But since , we know

$9-4i^2=9-4 \cdot (-1)=9+4=13$

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

7 + i

12 - 12i

24 + 7i

Explanation

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

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

$i\sqrt3 +7i$

$i\sqrt3 -7i$

$i\sqrt3-i$

$i\sqrt3 +i$

$2i\sqrt7$

Explanation

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

$i=\sqrt{-1}$

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

The correct answer is not listed.

Explanation

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

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:

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

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

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

$i\sqrt3 +7i$

$i\sqrt3 -7i$

$i\sqrt3-i$

$i\sqrt3 +i$

$2i\sqrt7$

Explanation

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

$i=\sqrt{-1}$

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

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

7 + i

12 - 12i

24 + 7i

Explanation

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

Define an operation $\Phi$ such that, for any complex number ,

$\Phi a = a + 2ai$

If $\Phi N = i$ , then evaluate .

$\frac{ 2}{5} + \frac{1}{5} i$

$\frac{ 2}{3} + \frac{1}{3} i$

$\frac{ 2}{5} - \frac{1}{5} i$

$\frac{ 1}{3} + \frac{2}{3} i$

$\frac{ 1}{5} - \frac{2}{5} i$

Explanation

$\Phi a = a + 2ai$ , so

$\Phi N = N + 2Ni = N (1+ 2i)$

$\Phi N = i$ , so

, and

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

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

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

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

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

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

$=\frac{ i - 2(-1)}{5}$

$=\frac{ i +2}{5}$

$=\frac{ 2}{5} + \frac{1}{5} i$

Define an operation $\Phi$ such that, for any complex number ,

$\Phi a = a + 2ai$

If $\Phi N = i$ , then evaluate .

$\frac{ 2}{5} + \frac{1}{5} i$

$\frac{ 2}{3} + \frac{1}{3} i$

$\frac{ 2}{5} - \frac{1}{5} i$

$\frac{ 1}{3} + \frac{2}{3} i$

$\frac{ 1}{5} - \frac{2}{5} i$

Explanation

$\Phi a = a + 2ai$ , so

$\Phi N = N + 2Ni = N (1+ 2i)$

$\Phi N = i$ , so

, and

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

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

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

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

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

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

$=\frac{ i - 2(-1)}{5}$

$=\frac{ i +2}{5}$

$=\frac{ 2}{5} + \frac{1}{5} i$

Simplify:

Explanation

Use FOIL:

Combine like terms:

But since , we know

$9-4i^2=9-4 \cdot (-1)=9+4=13$

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