Complex Imaginary Numbers - Algebra 2

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Question

Identify the real part of

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Answer

A complex number in its standard form is of the form: , where stands for the real part and stands for the imaginary part. The symbol stands for $\sqrt{-1}$ .

The real part in this problem is 1.

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Question

$\frac{-5+10i}{3+4i}=?$

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Answer

$\frac{-5+10i}{3+4i}$

$=\frac{(-5+10i)(3-4i)}{(3+4i)(3-4i)}$

$=\frac{-15+20i+30i-40i^{2}}{9-16i^{2}}$

$=\frac{-15+50i+40}{9+16}$

$=\frac{25+50i}{25}$

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Question

Multiply:

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Answer

Use the FOIL technique:

$= 7 \cdot 1 - 7 \cdot 2i + 3i \cdot 1 - 3i \cdot 2i$

$= 7 - 14i + 3i - 6i ^{2}$

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Question

Evaluate:

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

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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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Question

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

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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 the following product:

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Answer

Multiply these complex numbers out in the typical way:

${}(5+3i)(-2+i) = -10+5i-6i+3i^2$

and recall that by definition. Then, grouping like terms we get

which is our final answer.

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Question

Simplify:

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Answer

To add complex numbers, find the sum of the real terms, then find the sum of the imaginary terms.

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Question

Simplify:

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Answer

To add complex numbers, find the sum of the real terms, then find the sum of the imaginary terms.

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Question

Simplify:

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Answer

To add complex numbers, find the sum of the real terms, then find the sum of the imaginary terms.

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Question

Simplify:

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Answer

To subtract complex numbers, subtract the real terms together, then subtract the imaginary terms.

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Question

Simplify:

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Answer

To subtract complex numbers, subtract the real terms, then subtract the imaginary terms.

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Question

Simplify:

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Answer

To subtract complex numbers, subtract the real terms, then subtract the imaginary terms.

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Question

Simplify:

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Answer

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that $i=\sqrt-1$ , so .

Substitute in for

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Question

Simplify:

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Answer

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that $i=\sqrt-1$ , so .

Substitute in for

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Question

Simplify:

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Answer

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that $i=\sqrt-1$ , so .

Substitute in for

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Question

Simplify:

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Answer

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that $i=\sqrt{-1}$ , so .

Substitute in for .

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Question

Simplify:

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

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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:

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

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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

Write in standard form: $\frac{3}{2+4i}$

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Answer

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

Multiply by the conjugate:

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

$\frac{3(2-4i)}{4-16i^2}$

$\frac{6-12i}{4-16(-1)}$

$\frac{6-12i}{20}=\frac{6}{20}-\frac{12i}{20}=\mathbf{\frac{3}{10}-\frac{3i}{5}}$

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Question

Write in standard form: $\frac{12i}{2+4i}$

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Answer

$\frac{12i}{2+4i}$

Multiply by the conjugate:

$\frac{12i}{2+4i}*\frac{2-4i}{2-4i}$

$\frac{12i(2-4i)}{4-16i^2}$

Combine:

$\frac{24i-48i^2}{4-16i^2}$

Simplify:

$\frac{24i+48}{20}$

$\mathbf{\frac{6i}{5}+\frac{12}{5}}$

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