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SAT II Math I : Irrational Numbers

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

Example Question #21 : Number Theory

Simplify:

(2+3i)-(4-2i)

Possible Answers:

2+5i

-2+5i

6+i

-2+i

2-5i

Correct answer:

-2+5i

Explanation:

(2+3i)-(4-2i)

First remember the form of the complex number:

a+bi

Where "a" is the "Real" part, and "b" is the imaginary part.

Only the real parts can be combined together, and only the imaginary parts can be combined together.

Therefore the equation becomes after distributing the negative sign:

2+3i-4+2i

Collect the real and imaginary terms together:

2-4+3i+2i

Giving the answer:

-2+5i

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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:

$\small 0$

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

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

$\small -15$

$\small -15i$

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 #1 : Equations With Complex Numbers

Solve for and :

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

Possible Answers:

a=10, b=30

a=40, b=20

a=20, b=40

a=30, b=10

a=15, b=15

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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SAT II Math I : Irrational Numbers

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #21 : Number Theory

Example Question #1 : Equations With Complex Numbers

Example Question #1 : Equations With Complex Numbers

All SAT II Math I Resources

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