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GRE Subject Test: Math : Complex Conjugates

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Example Questions

Example Question #101 : Classifying Algebraic Functions

Evaluate $\small \frac{-6+2i}{10-3i}$

Possible Answers:

$\small \small \frac{-54+38i}{91-60i}$

$\small \small \frac{-66+2i}{91-60i}$

$\small \frac{-66+2i}{109}$

You cannot divide by complex numbers

$\small \frac{-54+38i}{109}$

Correct answer:

$\small \frac{-66+2i}{109}$

Explanation:

To divide by a complex number, we must transform the expression by multiplying it by the complex conjugate of the denominator over itself. In the problem, $\small 10-3i$ is our denominator, so we will multiply the expression by $\small \frac{10+3i}{10+3i}$ to obtain:

$\small \frac{-6+2i}{10-3i}\frac{10+3i}{10+3i}=\frac{-60+20i-18i+6i^2}{100-30i+30i-9i^2}$ .

We can then combine like terms and rewrite all $\small i^2$ terms as $\small -1$ . Therefore, the expression becomes:

$\small \frac{-60+2i+6i^2}{100-9i^2}=\frac{-66+2i}{109}$

Our final answer is therefore $\small \frac{-66+2i}{109}$

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Example Question #1 : Complex Conjugates

Simplify:

$\frac{5-2i}{4-i}$

Possible Answers:

6+3i

$\frac{22-3i}{17}$

$\frac{5i+6}{6}$

18-9i

Correct answer:

$\frac{22-3i}{17}$

Explanation:

To get rid of the fraction, multiply the numerator and denominator by the conjugate of the denominator.

$\frac{5-2i}{4-i}\left(\frac{4+i}{4+i}\right)$

Now, multiply and simplify.

$\frac{5-2i}{4-i}\left(\frac{4+i}{4+i}\right)=\frac{20+5i-8i-2i^2}{16-i^2}$

Remember that i^2=-1

$\frac{20+5i-8i-2i^2}{16-i^2}=\frac{20-3i-2(-1)}{16-(-1)}=\frac{22-3i}{17}$

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Example Question #1 : Complex Conjugates

Simplify:

$\frac{8-3i}{2-i}$

Possible Answers:

10-7i

$\frac{19+2i}{5}$

3i+6

$\frac{7+2i}{3}$

Correct answer:

$\frac{19+2i}{5}$

Explanation:

To get rid of the fraction, multiply the numerator and denominator by the conjugate of the denominator.

$\frac{8-3i}{2-i}\left(\frac{2+i}{2+i}\right)$

Now, multiply and simplify.

$\frac{8-3i}{2-i}\left(\frac{2+i}{2+i}\right)=\frac{16+8i-6i-3i^2}{4-i^2}$

Remember that i^2=-1

$\frac{16+8i-6i-3i^2}{4-i^2}=\frac{16+2i-3(-1)}{4-(-1)}=\frac{19+2i}{5}$

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Example Question #1 : Complex Conjugates

Divide: $\frac{2-i}{3+2i}$

The answer must be in standard form.

Possible Answers:

$\frac{2}{3}i$

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

$\frac{1}{5i}$

$\frac{4}{13}-\frac{7i}{13}$

Correct answer:

$\frac{4}{13}-\frac{7i}{13}$

Explanation:

Multiply both the numerator and the denominator by the conjugate of the denominator which is 3-2i which results in

$\frac{\left ( 2-i \right )\left ( 3-2i \right )}{\left ( 3+2i \right )\left ( 3-2i \right )}$

The numerator after simplification give us $6-4i-3i+2i^{2}=6-7i+2i^{2}=4-7i$

The denominator is equal to $3^{2}-4i^{2}=9+4=13$

Hence, the final answer in standard form =

$\frac{4}{13}-\frac{7i}{13}$

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Example Question #283 : Gre Subject Test: Math

$What\ is\ the\ complex\ conjugate\ for\ 3-i\sqrt{7}?$

Possible Answers:

$i\sqrt{7}$

$3+i\sqrt{7}$

$-i\sqrt{7}$

$-3-i\sqrt{7}$

Correct answer:

$3+i\sqrt{7}$

Explanation:

The definition of a complex conjugate is each of two complex numbers with the same real part and complex portions of opposite sign.

$In\ this\ case\ we\ have\ 3-i\sqrt{7}$

$to\ find\ its'\ complex\ conjugate\ we\ change\ the\ sign\ of\ the\ complex\ part$

$The\ complex\ conjugate\ is: 3+i\sqrt{7}$

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Example Question #4 : Complex Conjugates

Which of the following is the complex conjugate of (3+4i) ?

Possible Answers:

(3-4i)^3

(3-4i)

(3+4i)^2

Correct answer:

(3-4i)

Explanation:

The complex conjugate of a complex equation $(a+bi), \{a,b\}\in \mathbb {R}$ is (a-bi) .

The complex conjugate when multiplied by the original expression will also give me a real answer.

The complex conjugate of (3+4i) is (3-4i)

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Example Question #4 : Complex Conjugates

Simplify $\frac{14i+4}{7i-2}$

Possible Answers:

$\frac{20i-23}{24}$

$\frac{42i+99}{61}$

$\frac{3i-13}{9}$

$\frac{-91i+12}{15}$

$\frac{-90+56i}{-53}$

Correct answer:

$\frac{-90+56i}{-53}$

Explanation:

$\frac{14i+4}{7i-2}$

In problems like this, you are expected to simplify by removing i from the denominator. To do this, multiply the numerator and denominator by the conjugate of the denominator (switch the sign between the two terms from either a plus to a minus or vice versa) over itself. The conjugate over itself equals 1 and does not change the value of the expression (any number multiplied by 1 is still that number). Multiplying by the conjugate is the only way to eliminate i since there will be no middle term when we foil.

$\frac{14i+4}{7i-2}\cdot \frac{7i+2}{7i+2}$

$\frac{(14i+4)(7i+2)}{(7i-2)(7i+2)}$

$\frac{98i^{2}+28i+28i+8}{49i^{2}-4}$

Simplify i squared to -1 and then combine like terms

$\frac{98(-1)+28i+28i+8}{49(-1)-4}$

$\frac{-98+28i+28i+8}{-49-4}$

$\frac{-90+56i}{-53}$

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Example Question #3 : Complex Conjugates

Simplify $\frac{2+11i}{3-4i}$

Possible Answers:

$\frac{-38+41i}{25}$

$\frac{-9i-28}{11}$

$\frac{53i+8}{15}$

$\frac{30i+12}{15}$

$\frac{17+21i}{19}$

Correct answer:

$\frac{-38+41i}{25}$

Explanation:

$\frac{2+11i}{3-4i}$

$\frac{2+11i}{3-4i}\cdot \frac{3+4i}{3+4i}$

$\frac{(2+11i)(3+4i)}{(3-4i)(3+4i)}$

$\frac{6+33i+8i+44i^{2}}{9-16i^{2}}$

Simplify i squared to be -1 and then combine like terms

$\frac{6+33i+8i+44(-1)}{9-16(-1)}$

$\frac{6+33i+8i-44}{9+16}$

$\frac{-38+41i}{25}$

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Example Question #102 : Classifying Algebraic Functions

Simplify $\frac{7i}{-5+4i}$

Possible Answers:

$\frac{-11i-24}{9}$

$\frac{22i-14}{17}$

$\frac{83i+46}{25}$

$\frac{-35i+28}{41}$

$\frac{42i-17}{45}$

Correct answer:

$\frac{-35i+28}{41}$

Explanation:

$\frac{7i}{-5+4i}$

$\frac{7i}{-5+4i}\cdot \frac{-5-4i}{-5-4i}$

$\frac{7i(-5-4i)}{(-5+4i)(-5-4i)}$

$\frac{-35i-28i^{2}}{25-16i^{2}}$

Simplify i squared to be -1 and then combine like terms

$\frac{-35i-28(-1)}{25-16(-1)}$

$\frac{-35i+28}{25+16}$

$\frac{-35i+28}{41}$

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Example Question #4 : Complex Conjugates

Simplify $\frac{8i-2}{4i+6}$

Possible Answers:

$\frac{-16i-19}{13}$

$\frac{-63i+14}{12}$

$\frac{-20-56i}{-52}$

$\frac{32i+19}{34}$

$\frac{-5-14i}{-13}$

Correct answer:

$\frac{-5-14i}{-13}$

Explanation:

$\frac{8i-2}{4i+6}$

$\frac{8i-2}{4i+6}\cdot \frac{4i-6}{4i-6}$

$\frac{(8i-2)(4i-6)}{(4i+6)(4i-6)}$

$\frac{32i^{2}-8i-48i+12}{16i^{2}-36}$

Simplify i squared to be -1 and then combine like terms

$\frac{32(-1)-8i-48i+12}{16(-1)-36}$

$\frac{-32-8i-48i+12}{-16-36}$

$\frac{-20-56i}{-52}$

The coefficients of all the terms can divide by 4 so reduce each of them

$\frac{-5-14i}{-13}$

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GRE Subject Test: Math : Complex Conjugates

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #101 : Classifying Algebraic Functions

Example Question #1 : Complex Conjugates

Example Question #1 : Complex Conjugates

Example Question #1 : Complex Conjugates

Example Question #283 : Gre Subject Test: Math

Example Question #4 : Complex Conjugates

Example Question #4 : Complex Conjugates

Example Question #3 : Complex Conjugates

Example Question #102 : Classifying Algebraic Functions

Example Question #4 : Complex Conjugates

All GRE Subject Test: Math Resources

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