GRE Subject Test: Math : Classifying Algebraic Functions

Study concepts, example questions & explanations for GRE Subject Test: Math

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Example Questions

Example Question #2 : Operations On Complex Numbers

\(\displaystyle Simplify: (13-7i)-(10-8i)\)

 

Possible Answers:

\(\displaystyle 23-i\)

\(\displaystyle 3+i\)

\(\displaystyle 3-15i\)

\(\displaystyle -74-174i\)

Correct answer:

\(\displaystyle 3+i\)

Explanation:

\(\displaystyle When\ subtracting\ binomials\ in\ ( )\; first\ distribute\ the\ negative\ sign.\)

\(\displaystyle (13-7i)-(10-8i)=13-7i-10+8i\)

\(\displaystyle Now\ combine\ like\ terms.\)

\(\displaystyle 13-10-7i+8i=3+i\)

Example Question #2 : Operations On Complex Numbers

What is the value of \(\displaystyle (3-4i)(2i+3)\)?

Possible Answers:

\(\displaystyle -6i-17\)

None of the other answers

\(\displaystyle -6i+17\)

\(\displaystyle -6i+1\)

\(\displaystyle 6i-17\)

Correct answer:

\(\displaystyle -6i+17\)

Explanation:

Distribute and Multiply:

\(\displaystyle (3-4i)(2i+3)=6i+9-8i^2-12i\)

Simplify all terms...

\(\displaystyle 6i+9-8(-1)-12i\)

\(\displaystyle =6i+9+8-12i\)

\(\displaystyle =-6i+17\)

Example Question #5 : Operations On Complex Numbers

What is the value: \(\displaystyle i^{24}(i^{3}+i^{-2})\)?

Possible Answers:

\(\displaystyle -i-1\)

\(\displaystyle 1+i\)

\(\displaystyle -1-i\)

\(\displaystyle -(-i+1)\)

Correct answer:

\(\displaystyle -1-i\)

Explanation:

Step 1: Recall the cycle of imaginary numbers to a random power \(\displaystyle x\).

If \(\displaystyle x=1\), then \(\displaystyle i^1=i\)

If \(\displaystyle x=2\), then \(\displaystyle i^2=-1\)

If \(\displaystyle x=3\), then \(\displaystyle i^3=-i\)

If \(\displaystyle x=4\), then \(\displaystyle i^4=1\)

If \(\displaystyle x=5\), then \(\displaystyle i^5=i^4(i^1)=1(i)=i\)

and so on....

The cycle repeats every \(\displaystyle 4\) terms. 

For ANY number \(\displaystyle x\ge 5\), you can break down that term into smaller elementary powers of i. 

Step 2: Distribute the \(\displaystyle i^{24}\) to all terms in the parentheses:

\(\displaystyle (i^{24})(i^3)+(i^{24})(i^{-2})\).

Step 3: Recall the rules for exponents:

\(\displaystyle \forall\{a,b,c,d\} \in \mathbb{R}::\)

\(\displaystyle (a^b)\cdot (a^c)=a^{b+c}\)

\(\displaystyle {a^{-b}}=\frac {1}{a^b}\)

\(\displaystyle (a^b)\cdot (a^{-c})=a^b \cdot \bigg (\frac {1}{a^c}\bigg)=\frac {a^b}{a^c}=a^{b-c}=a^d\)

Step 4: Use the rules to rewrite the expression in Step 2:

\(\displaystyle (i^{24})\cdot {i^3}=i^{24+3}=i^{27}\)

\(\displaystyle (i^{24})(i^{-2})=\frac {i^{24}}{i^2}=i^{24-2}=i^{22}\)

Step 5: Simplify the results in Step 4. Use the rules in Step 1.:

\(\displaystyle i^{27}=(i^{24}\cdot i^3)=1(-i)=-i\)

\(\displaystyle i^{22}=(i^{20}\cdot i^{2})=1(-1)=-1\)

Step 6: Write the answer in \(\displaystyle a+bi\) form, where \(\displaystyle a\) is the real part and \(\displaystyle bi\) is the imaginary part:

We get \(\displaystyle -1-i\)

Example Question #3 : Operations On Complex Numbers

\(\displaystyle Add: (3-5i)+(2+7i)\)

Possible Answers:

\(\displaystyle 1-12i\)

\(\displaystyle 1+12i\)

\(\displaystyle 5+2i\)

\(\displaystyle 6-35i\)

Correct answer:

\(\displaystyle 5+2i\)

Explanation:

When adding complex numbers, we add the real numbers and add the imaginary numbers. 

\(\displaystyle Add: (3-5i)+(2+7i)\)

\(\displaystyle =(3+2)+(-5i+7i)\)

\(\displaystyle =5+2i\)

 

 

Example Question #4 : Operations On Complex Numbers

\(\displaystyle Subtract: (15+7i)-(12+9i)\)

Possible Answers:

\(\displaystyle 3-2i\)

\(\displaystyle 3+16i\)

\(\displaystyle 3-5i\)

\(\displaystyle 3-16i\)

Correct answer:

\(\displaystyle 3-2i\)

Explanation:

In order to subtract complex numbers, we must first distribute the negative sign to the second complex number. 

\(\displaystyle Subtract: (15+7i)-(12+9i)\)

\(\displaystyle 15+7i-12-9i\)

\(\displaystyle 15-12+7i-9i=3-2i\)

 

Example Question #4 : Operations On Complex Numbers

\(\displaystyle Simplify: 3i(-5+2i)\)

Possible Answers:

\(\displaystyle 15+6i^2\)

\(\displaystyle -6-15i\)

\(\displaystyle 15-6i\)

\(\displaystyle 6i^2-15i\)

Correct answer:

\(\displaystyle -6-15i\)

Explanation:

\(\displaystyle Simplify: 3i(-5+2i)\)

First we must distribute

\(\displaystyle 3i(-5+2i)=-15i+2i^2\)

\(\displaystyle We\ must\ remember\ that\ i^2=-1\)

\(\displaystyle -15i+6i^2=-15i+6(-1)\)

\(\displaystyle -15i+6(-1)=-6-15i\ (i\ term\ is\ always\ last)\)

 

 

 

Example Question #21 : Imaginary Numbers & Complex Functions

\(\displaystyle Simplify: (4-7i)(1-2i)\)

Possible Answers:

\(\displaystyle 4-14i\)

\(\displaystyle -10-15i\)

\(\displaystyle 5-9i\)

\(\displaystyle 4-15i+14i^2\)

Correct answer:

\(\displaystyle -10-15i\)

Explanation:

\(\displaystyle In\ order\ to\ simplify\ (4-7i)(1-2i)\ we\ must\ multiply\ the\ binomials\ using\ the\ FOIL\ method.\)

\(\displaystyle F\ is\ for\ first\ times\ first: =4*1=4\)

\(\displaystyle O\ is\ for\ outside\ times\ outside =4*(-2i)=-8i\)

\(\displaystyle I\ is\ for\ inside\ times\ inside=(-7i)*1=-7i\)

\(\displaystyle L\ is\ for\ last\ times\ last =(-7i)*(-2i)=14i^2\)

\(\displaystyle Now\ we\ combine\ like\ terms\ and\ simplify:\)

\(\displaystyle 4-8i-7i+14i^2\)

\(\displaystyle ^* Remember\ that\ i^2=-1\)

\(\displaystyle 4-15i+14(-1)=-10-15i\)

Example Question #81 : Classifying Algebraic Functions

\(\displaystyle Simplify: (5-8i)^2\)

Possible Answers:

\(\displaystyle 25-64i\)

\(\displaystyle 25-64i^2\)

\(\displaystyle -39\)

\(\displaystyle -39-80i\)

Correct answer:

\(\displaystyle -39-80i\)

Explanation:

\(\displaystyle When\ a\ binomial\ is\ squared\ we\ must\ write\ it\ out\ twice\ and\ perform\ the\ FOIL\ method.\)

\(\displaystyle (5-8i)^2=(5-8i)(5-8i)\)

\(\displaystyle F\ is\ for\ first\ times\ first = 5*5=25\)

\(\displaystyle O\ is\ for\ outside\ times\ outside = 5*8i=40i\)

\(\displaystyle I\ is\ for\ inside\ times\ inside: 8i*5=40i\)

\(\displaystyle L\ is\ for\ last\ times\ last = 8i*8i=64i^2=64(-1)=-64\)

Now we put each of these together and combine like terms: 

\(\displaystyle 25-40i-40i-64=-39-80i\)

Example Question #82 : Classifying Algebraic Functions

\(\displaystyle \frac{\sqrt{-16}}{\sqrt{-8}}\)

Possible Answers:

\(\displaystyle -\sqrt{2}\)

\(\displaystyle 2\)

\(\displaystyle 4\)

\(\displaystyle \sqrt{2}\)

\(\displaystyle -2\)

Correct answer:

\(\displaystyle \sqrt{2}\)

Explanation:

\(\displaystyle \frac{\sqrt{-16}}{\sqrt{-8}}\)

Take i (the square root of -1) out of both radicals then divide.

\(\displaystyle \frac{i\sqrt{16}}{i\sqrt{8}}\)

\(\displaystyle \frac{\sqrt{16}}{\sqrt{8}}\)

\(\displaystyle \sqrt{2}\)

Example Question #271 : Gre Subject Test: Math

\(\displaystyle \sqrt{-16}\cdot \sqrt{-25}\)

Possible Answers:

\(\displaystyle -20\)

\(\displaystyle 20\)

\(\displaystyle 40i\)

\(\displaystyle 4\sqrt{5}\)

\(\displaystyle 20i\)

Correct answer:

\(\displaystyle -20\)

Explanation:

\(\displaystyle \sqrt{-16}\cdot \sqrt{-25}\)

Take out i (the square root of -1) from both radicals and then multiply. You are not allowed to first multiply the radicals and then simplify because the roots are negative.

\(\displaystyle 4i\cdot 5i\)

\(\displaystyle 20i^{2}\)

Make i squared -1

\(\displaystyle 20(-1)\)

\(\displaystyle -20\)

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