GRE Quantitative Reasoning - Complex Conjugates

1

Simplify

$\frac{5+2i}{-3-2i}$

$\frac{-19+4i}{13}$

$\frac{16+4i}{3}$

$\frac{i+2}{10}$

$\frac{7i+1}{5}$

Explanation

$\frac{5+2i}{-3-2i}$

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.

$\frac{5+2i}{-3-2i}\cdot \frac{-3+2i}{-3+2i}$

$\frac{(5+2i)(-3+2i)}{(-3-2i)(-3+2i)}$

$\frac{-15-6i+10i+4i^{2}}{9-4i^{2}}$

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

$\frac{-15-6i+10i+4(-1)}{9-4(-1)}$

$\frac{-15+4i-4}{9+4}$

$\frac{-19+4i}{13}$

2

Which of the following is the complex conjugate of ?

Explanation

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

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

The complex conjugate of is

3

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

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

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

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

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

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

Explanation

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

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.

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

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

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

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

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

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

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

4

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

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

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

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

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

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

Explanation

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

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.

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

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

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

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

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

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

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

5

Simplify:

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

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

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

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

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

6

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

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

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

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

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

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

Explanation

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

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.

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

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

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

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

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

7

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

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

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

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

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

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

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

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

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

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

8

Simplify $\frac{2i}{9i+3}$

$\frac{3+i}{15}$

$\frac{-18-6i}{-90}$

$\frac{23i-6}{72}$

$\frac{-38+2i}{13}$

$\frac{49i+27}{52}$

Explanation

$\frac{2i}{9i+3}$

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.

$\frac{2i}{9i+3}\cdot \frac{9i-3}{9i-3}$

$\frac{2i(9i-3)}{(9i+3)(9i-3)}$

$\frac{18i^{2}-6i}{81i^{2}-9}$

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

$\frac{18(-1)-6i}{81(-1)-9}$

$\frac{-18-6i}{-81-9}$

$\frac{-18-6i}{-90}$

Since each term divides by a greatest common factor of -6 reduce all of the coefficients. It would also be equivalent to divide by 6 to reduce all of the terms.

$\frac{3+i}{15}$

9

Simplify $\frac{-19i-8}{-5i-2}$

$\frac{-111+2i}{-29}$

$\frac{47+78i}{21}$

$\frac{-30i+26}{13}$

$\frac{-78i+49}{5}$

$\frac{10i-3}{29}$

Explanation

$\frac{-19i-8}{-5i-2}$

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.

$\frac{-19i-8}{-5i-2}\cdot \frac{-5i+2}{-5i+2}$

$\frac{(-19i-8)(-5i+2)}{(-5i-2)(-5i+2)}$

$\frac{95i^{2}+40i-38i-16}{25i^{2}-4}$

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

$\frac{95(-1)+40i-38i-16}{25(-1)-4}$

$\frac{-95+40i-38i-16}{-25-4}$

$\frac{-111+2i}{-29}$

10

Simplify $\frac{14i-9}{9+i}$

$\frac{135i-67}{82}$

$\frac{14i+111}{42}$

$\frac{23i-11}{65}$

$\frac{-44i+63}{86}$

$\frac{54i-18}{25}$

Explanation

$\frac{14i-9}{9+i}$

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.

$\frac{14i-9}{9+i}\cdot \frac{9-i}{9-i}$