Operations on Complex Numbers - GRE Subject Test: Math

Card 0 of 20

Question

Expand and Simplify:

Answer

Step 1: We will multiply the two complex conjugates: and .

Step 2: Replace with .

Simplify:

Step 3: Multiply the result of the complex conjugates to the other parentheses,.

The final answer after the product of all three binomials is

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Question

Expand: $(i-1)^{10}$ .

Answer

Quick Way:
Step 1: Expand .

.

Remember:

Step 2: $(1-i)^{10}=(i-1)^{2\cdot 5}$

By this equivalence, I can just raise the answer of to the power .

$(i-1)^{10}=(-2i)^{5}=-32i^5$

$i^5 \equiv i^1=i$ . Replace ..

Final answer:

Long Way:

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Question

Multiply:

Answer

Step 1: FOIL:

Recall, FOIL means to multiply the first terms in both binomials together, the outer terms together, the inner terms together, and finally, the last terms together.

Step 2: Simplify:

Step 3: Recall: . Replace and simplify.

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Question

What is the value of ?

Answer

Distribute and Multiply:

Simplify all terms...

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Question

Answer

When adding imaginary numbers, simply add the real parts and the imaginary parts.

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Question

Answer

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

$Now\ combine\ like\ terms.$

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Question

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

Answer

Step 1: Recall the cycle of imaginary numbers to a random power .

If , then

If , then

If , then

If , then

If , then

and so on....

The cycle repeats every terms.

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

Step 2: Distribute the $i^{24}$ to all terms in the parentheses:

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

Step 3: Recall the rules for exponents:

$\forall{a,b,c,d} \in \mathbb{R}::$

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

${a^{-b}}=\frac {1}{a^b}$

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

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

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

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

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

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

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

Step 6: Write the answer in form, where is the real part and is the imaginary part:

We get

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Question

Answer

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

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Question

Answer

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

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Question

Answer

First we must distribute

$We\ must\ remember\ that\ i^2=-1$

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

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Question

Answer

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

$F\ is\ for\ first\ times\ first: =41=4$

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

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

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

$Now\ we\ combine\ like\ terms\ and\ simplify:$

$^* Remember\ that\ i^2=-1$

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Question

Answer

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

$F\ is\ for\ first\ times\ first = 55=25$

$O\ is\ for\ outside\ times\ outside = 58i=40i$

$I\ is\ for\ inside\ times\ inside: 8i5=40i$

$L\ is\ for\ last\ times\ last = 8i8i=64i^2=64(-1)=-64$

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

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Question

$\sqrt{-7}\cdot \sqrt{-4}$

Answer

$\sqrt{-7}\cdot \sqrt{-4}$

First, 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.

$i\sqrt{7}\cdot 2i$

$2i^{2}\sqrt{7}$

Change i squared to -1

$2(-1)\sqrt{7}$

$-2\sqrt{7}$

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Question

$\sqrt{-16}\cdot \sqrt{-25}$

Answer

$\sqrt{-16}\cdot \sqrt{-25}$

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.

$4i\cdot 5i$

$20i^{2}$

Make i squared -1

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Question

$\frac{\sqrt{-16}}{\sqrt{-8}}$

Answer

$\frac{\sqrt{-16}}{\sqrt{-8}}$

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

$\frac{i\sqrt{16}}{i\sqrt{8}}$

$\frac{\sqrt{16}}{\sqrt{8}}$

$\sqrt{2}$

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Question

$\frac{\sqrt{-51}}{\sqrt{3}}$

Answer

$\frac{\sqrt{-51}}{\sqrt{3}}$

Take i (the square root of -1) out of the radical.

$\frac{i\sqrt{51}}{\sqrt{3}}$

$i\sqrt{17}$

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Question

$\sqrt{5}\cdot \sqrt{-42}$

Answer

$\sqrt{5}\cdot \sqrt{-42}$

Take out i (the square root of -1) from the radical and then multiply.

$\sqrt{5}\cdot i\sqrt{42}$

$i\sqrt{210}$

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Question

$\sqrt{-45}\cdot \sqrt{72}$

Answer

$\sqrt{-45}\cdot \sqrt{72}$

Take out i (the square root of -1) and then simplify before multiplying.

$i\sqrt{45}\cdot \sqrt{72}$

$i\sqrt{9\cdot 5}\cdot \sqrt{36\cdot 2}$

$3i\sqrt{5}\cdot 6\sqrt{2}$

$18i\sqrt{10}$

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Question

$-\sqrt{-16}\cdot \sqrt{-40}$

Answer

$-\sqrt{-16}\cdot \sqrt{-40}$

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.

$-i\sqrt{16}\cdot i\sqrt{40}$

$-4i\cdot i\sqrt{4\cdot 10}$

$-4i\cdot 2i\sqrt{10}$

$-8i^{2}\sqrt{10}$

$-8(-1)\sqrt{10}$

$8\sqrt{10}$

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Question

$-\sqrt{-20}\cdot -\sqrt{32}$

Answer

$-\sqrt{-20}\cdot -\sqrt{32}$

Take out i (the square root of -1) from the radical, simplify, and then multiply.

$-i\sqrt{20}\cdot -\sqrt{32}$

$-i\sqrt{5\cdot 4}\cdot -\sqrt{16\cdot 2}$

$-2i\sqrt{5}\cdot -4\sqrt{2}$

$8i\sqrt{10}$

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