Imaginary Numbers & Complex Functions - GRE Subject Test: Math

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Question

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

The answer must be in standard form.

Answer

Multiply both the numerator and the denominator by the conjugate of the denominator which is 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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Question

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

Answer

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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Question

Simplify:

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

Answer

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

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

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Question

Simplify:

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

Answer

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

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

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Question

Which of the following is the complex conjugate of ?

Answer

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

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Question

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

Answer

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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Question

Simplify

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

Answer

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

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

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

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

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

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Question

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

Answer

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

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

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

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

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

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Question

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

Answer

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

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

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

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

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

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Question

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

Answer

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

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

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

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Question

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

Answer

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

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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Question

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

Answer

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

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.

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

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Question

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

Answer

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

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

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

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

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

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Question

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

Answer

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

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.

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

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

$\frac{126i-81-14i^{2}+9i}{81-i^{2}}$

Simplify i squared to be -1 and combine like terms

$\frac{126i-81-14(-1)+9i}{81-(-1)}$

$\frac{126i-81+14+9i}{81+1}$

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

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Question

Simplify $\frac{13i+13}{13i-9}$

Answer

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

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.

$\frac{13i+13}{13i-9}\cdot \frac{13i+9}{13i+9}$

$\frac{(13i+13)(13i+9)}{(13i-9)(13i+9)}$

$\frac{(169i^{2}+169i+117i+117)}{(169i^{2}-81)}$

Simplify i squared to be -1 and combine like terms

$\frac{(169(-1)+169i+117i+117)}{(169)(-1)-81}$

$\frac{(-169+169i+117i+117)}{(-169-81)}$

$\frac{-52+286i}{-250}$

Each term divides by 2 so make sure to reduce all of the terms

$\frac{-26+143i}{-125}$

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Question

Simplify $\frac{-12i-17}{6i+11}$

Answer

$\frac{-12i-17}{6i+11}$

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.

$\frac{-12i-17}{6i+11}\cdot \frac{6i-11}{6i-11}$

$\frac{(-12i-17)(6i-11)}{(6i+11)(6i-11)}$

$\frac{-72i^{2}+121i-102i+187}{36i^{2}-121}$

Simplify i squared to be -1 and combine like terms

$\frac{-72(-1)+121i-102i+187}{36(-1)-121}$

$\frac{72+121i-102i+187}{-36-121}$

$\frac{19i+259}{-157}$

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Question

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

Answer

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

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.

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

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

$\frac{3i+i^{2}-6-2i}{9-i^{2}}$

Simplify i squared to be -1 and combine like terms

$\frac{3i+(-1)-6-2i}{9-(-1)}$

$\frac{3i-1-6-2i}{9+1}$

$\frac{i-7}{10}$

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Question

Simplify $\frac{3i+7}{4i-4}$

Answer

$\frac{3i+7}{4i-4}$

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.

$\frac{3i+7}{4i-4}\cdot \frac{4i+4}{4i+4}$

$\frac{(3i+7)(4i+4)}{(4i-4)(4i+4)}$

$\frac{12i^{2}+12i+28i+28}{16i^{2}-16}$

Simplify i squared to be -1 and combine like terms

$\frac{12(-1)+12i+28i+28}{16(-1)-16}$

$\frac{-12+12i+28i+28}{-16-16}$

$\frac{40i+16}{-32}$

Since every term divides by 8 make sure to reduce all the terms by that greatest common factor

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

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Question

Write the complex number $1-\sqrt{3}i$ in polar form, where polar form expresses the result in terms of a distance from the origin on the complex plane and an angle from the positive -axis, $\theta$ , measured in radians.

Answer

To see what the polar form of the number is, it helps to draw it on a graph, where the horizontal axis is the imaginary part and the vertical axis the real part. This is called the complex plane.

To find the angle $\theta$ , we can find its supplementary angle $\Psi$ and subtract it from $\pi$ radians, so $\theta = \pi - \Psi$ .

Using trigonometric ratios, $\tan \Psi = \frac{1}{\sqrt{3}}$ and $\arctan({\frac{1}{\sqrt{3}}})= \Psi = \frac{\pi}{6}$ .

Then $\theta = \frac{5\pi}{6}$ .

To find the distance , we need to find the distance from the origin to the point $(-\sqrt{3}, 1)$ . Using the Pythagorean Theorem to find the hypotenuse , $r = \sqrt{ \sqrt{3}^2 + 1^2 }$ or .

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Question

Evaluate:

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

Answer

We can set in the cube of a binomial pattern:

$\left ( A + B\right ) ^{3} = A ^{3} + 3 A^{2 }B + 3AB^{2} + B^{3}$

$= 2 ^{3} + 3 \cdot 2 ^{2 } \cdot 3i + 3 \cdot 2 \cdot (3i)^{2} + (3i)^{3}$

$= 2^{3} + 3 \cdot 2^{2} \cdot 3i + 3 \cdot 2 \cdot 3 ^{2}\cdot i^{2} + 3^{3} \cdot i^{3}$

$= 8 + 36i + 54 i^{2} + 27 i^{3}$

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