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Algebra II : Equations with Complex Numbers

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Example Questions

Example Question #1 : Equations With Complex Numbers

x, y are real numbers.

$4x+ \left (3y+ 4 \right ) i = 21 + 7i$

Evaluate .

Possible Answers:

y = 1

y = 6.25

y = -1

y =- 5.25

y = 5.25

Correct answer:

y = 1

Explanation:

For two imaginary numbers to be equal to each other, their imaginary parts must be equal. Therefore, we set, and solve for in:

3y + 4 = 7

3y = 3

y = 1

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Example Question #1 : Equations With Complex Numbers

If $\small x$ and $\small y$ are real numbers, and $\small \small 3x+y+(2x+2y)i=13+10i$ , what is $\small z$ if $\small z=x-4y$ ?

Possible Answers:

$\small -15$

$\small 0$

$\small -15i$

$\small -16\frac{1}{4}$

$\small -13\frac{3}{4}$

Correct answer:

$\small 0$

Explanation:

To solve for $\small z$ , we must first solve the equation with the complex number for $\small x$ and $\small y$ . We therefore need to match up the real portion of the compex number with the real portions of the expression, and the imaginary portion of the complex number with the imaginary portion of the expression. We therefore obtain:

$\small 3x+y=13$ and $\small 2x+2y=10$

We can use substitution by noticing the first equation can be rewritten as $\small y=13-3x$ and substituting it into the second equation. We can therefore solve for $\small x$ :

$\small 2x+2(13-3x)=10$

$\small 2x+26-6x=10$

$\small -4x=-16$

$\small x=4$

With this $\small x$ value, we can solve for $\small y$ :

$\small 3(4)+y=13$

$\small 12+y=13$

$\small y=1$

Since we now have $\small x$ and $\small y$ , we can solve for $\small z$ :

$\small z=x-4y$

$\small \small z=(4)-4(1)=0$

Our final answer is therefore $\small z=0$

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Example Question #2 : Equations With Complex Numbers

Solve for if $x = 4\cdot \left ( i^2 + 3\right )$ .

Possible Answers:

Correct answer:

Explanation:

Go about this problem just like any other algebra problem by following your order of operations. We will first evaluate what is inside the parentheses: i^2 + 3 . At this point, we need to know the properties of which are as follows:

$i = \sqrt{-1}$

i^2 = -1

$i^3 = -\sqrt{-1}$

i^4 = 1

Therefore, i^2 + 3 = -1+3 = 2 and the original expression becomes $x = 4 \cdot (2) = 8$

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Example Question #1 : Equations With Complex Numbers

Evaluate and simplify (i+3)^2 .

Possible Answers:

8+6i

None of the other answers.

9+6i

Correct answer:

8+6i

Explanation:

The first step is to evaluate the expression. By FOILing the expression, we get:

(i+3)(i+3)

$=i\cdot i+3i+3i+3\cdot 3$

=i^2 + 6i + 9

Now we need to simplify any terms that we can by using the properties of

$i = \sqrt-1$

i^2 = -1

$i^3 = -\sqrt-1$

i^4 = 1

Therefore, the expression becomes

-1 + 6i + 9 = 8+6i

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Example Question #5 : Equations With Complex Numbers

Solve for :

(6+3i)^2=4i-30z

Possible Answers:

$\frac{-27+32i}{30}$

$\frac{-45-32i}{30}$

$\frac{27-32i}{30}$

$\frac{27+32i}{30}$

$\frac{-27-32i}{30}$

Correct answer:

$\frac{-27-32i}{30}$

Explanation:

In order to solve this problem, we need to first simplify our equation. The first thing we should do is distribute the square, which gives us

(6+3i)(6+3i)=36+18i+18i+9i^2=36+36i+9i^2

Now i^2 is actually just . Therefore, this becomes

36+36i-9=27+36i

Now all we need to do is solve for in the equation:

$27+\underset{-4i}{36i}=\underset{-4i}{4i}-30z$

which gives us

27+32i=30z

Finally, we get

$\frac{27+32i}{-30}=\frac{-30z}{-30}$

and therefore, our solution is

$z = \frac{-27-32i}{30}$

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Example Question #1 : Equations With Complex Numbers

Solve for and :

$ai^{93}+bi^{35}+ai^{24}-bi^{86}=40+20i$

Possible Answers:

a=30, b=10

a=15, b=15

a=40, b=20

a=20, b=40

a=10, b=30

Correct answer:

a=30, b=10

Explanation:

Remember that

$i = \sqrt{-1}\\ i^2 = \sqrt{-1}^2 = -1\\ i^3 = i^2\cdot i = -1 \cdot i = -i\\ i^4 = i^2 \cdot i^2 = (-1)\cdot(-1) = 1\\ i^5 = i^4\cdot i = 1\cdot i = i = \sqrt{-1}\\ i^6 = i^4\cdot i^2 = 1\cdot i^2 = i^2 =-1$

So the powers of are cyclic. This means that when we try to figure out the value of an exponent of , we can ignore all the powers that are multiples of because they end up multiplying the end result by , and therefore do nothing.

This means that

$ai^{93} + bi^{35} + ai^{24} - bi^{86}\\ = ai^{92+1} + bi^{32+3} + ai^{24+0} - bi^{84+2}\\ = ai^{92}\cdot i^1 + bi^{32}\cdot i^3 + ai^{24}\cdot i^0 - bi^{84}\cdot i^2\\ =a(i^4)^{23}\cdot i^1 + b(i^4)^8\cdot i^3 + a(i^4)^6\cdot i^0 - b(i^4)^{21}\cdot i^2\\ = a (1^{23})\cdot i + b(1^8)\cdot i^3 + a(1^6) - b(1^{21})\cdot i^2$

Now, remembering the relationships of the exponents of , we can simplify this to:

$ai - bi + a - (-b) = ai-bi+a+b\\ = (a+b) + (a-b)i = 40+20i$

Because the elements on the left and right have to correspond (no mixing and matching!), we get the relationships:

a+b=40

a-b=10

No matter how you solve it, you get the values a=30 , b=10 .

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Example Question #1 : Equations With Complex Numbers

Solve

$2x^{^{2}} - 1 = 2x^{2} +3$

Possible Answers:

x=1

x=3

x=2

No solution

All real numbers

Correct answer:

No solution

Explanation:

To solve

$2x^{^{2}} - 1 = 2x^{2} +3$

Subtract $2x^{2}$ from both side:

-1=3

Which is never true, so there is no solution.

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Algebra II : Equations with Complex Numbers

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Equations With Complex Numbers

Example Question #1 : Equations With Complex Numbers

Example Question #2 : Equations With Complex Numbers

Example Question #1 : Equations With Complex Numbers

Example Question #5 : Equations With Complex Numbers

Example Question #1 : Equations With Complex Numbers

Example Question #1 : Equations With Complex Numbers

All Algebra II Resources

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