SAT Math - How to multiply complex numbers

$x^{4}-1$ 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:

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

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

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:

Explanation

Use FOIL:

Combine like terms:

But since , we know

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

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

None of the other choices gives the correct response.

Explanation

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:

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

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

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

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

$90- 101i \sqrt{7}$

$90- 115i\sqrt{7}$

$216 - 7i\sqrt{7}$

$342- 115i\sqrt{7}$

$342- 101i \sqrt{7}$

Explanation

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:

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

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

Remember that $i= \sqrt{-1}$ .

Simplify:

Explanation

Use FOIL to multiply complex numbers as follows:

Since $i= \sqrt{-1}$ , it follows that , so then:

$2-2i+3i-3 \cdot (-1)$

Combining like terms gives:

Raise to the power of 3.

None of the other choices gives the correct response.

Explanation

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 :

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

Taking advantage of the Power of a Product Rule:

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

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

and

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

$(6-7i)^{3} =216 - 3\cdot 36 \cdot 7i + 3 \cdot 6 \cdot 49 \cdot (-1) - 343 \cdot (-i)$

$(6-7i)^{3} =216 - 756 i -882 + 343 i$

Collecting real and imaginary terms:

$(6-7i)^{3} =216 -882 - 756 i + 343 i$

$(6-7i)^{3} =-666 - 413 i$

; is the complex conjugate of .

Evaluate

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

Explanation

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

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

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

None of the other choices gives the correct response.

$-108,000 i \sqrt{5}$

$-108,000 \sqrt{5}$

Explanation

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:

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

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

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

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

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

This is not among the given choices.

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

Evaluate

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

$\textup{None of the other choices gives the correct response}$

Explanation

$x^{2} - 2xy + y^{2}$ 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 $i^{2} = -1$ :

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

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

How to multiply complex numbers

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