GRE Quantitative Reasoning - Operations on Complex Numbers

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

$-2\sqrt{7}$

$\sqrt{28}$

$i\sqrt{28}$

$2\sqrt{7}$

$-i\sqrt{7}$

Explanation

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

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

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

$-2\sqrt{7}$

Expand and Simplify:

Explanation

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

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

Explanation

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

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

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

$4\sqrt{5}$

Explanation

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

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

Explanation

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

Explanation

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

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

$\sqrt{2}$

$-\sqrt{2}$

Explanation

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

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

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

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

$\sqrt{2}$

Multiply:

Explanation

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.