Complex Numbers

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ACT Math › Complex Numbers

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Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Which of the following is incorrect?

Explanation

Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Thus, to balance the equation, add like terms on the left side.

$1+5i \neq 1+9i$

Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Which of the following is incorrect?

Explanation

Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Thus, to balance the equation, add like terms on the left side.

$1+5i \neq 1+9i$

Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Simplify: $\frac{6+12i}{4}$

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

$\frac{3}{2} + 12i$

$\frac{3}{2} + \frac{12i}{4i}$

$\textup{None of these answers are correct}$

Explanation

This problem can be solved very similarly to a binomial such as . In this case, both the real and nonreal terms in the complex number are eligible to be divided by the real divisor.

$\frac{6+12i}{4} = (\frac{6}{4} + \frac{12i}{4})$

$\frac{12i}{4} = \frac{12}{4} \cdot i = 3 \cdot i = 3i$ , so

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

Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Simplify: $\frac{6+12i}{4}$

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

$\frac{3}{2} + 12i$

$\frac{3}{2} + \frac{12i}{4i}$

$\textup{None of these answers are correct}$

Explanation

This problem can be solved very similarly to a binomial such as . In this case, both the real and nonreal terms in the complex number are eligible to be divided by the real divisor.

$\frac{6+12i}{4} = (\frac{6}{4} + \frac{12i}{4})$

$\frac{12i}{4} = \frac{12}{4} \cdot i = 3 \cdot i = 3i$ , so

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

Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Which of the following isincorrect?

Explanation

A problem like this can be solved similarly to a linear binomial like /

$-3 +8i \neq 3+8i$

Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Which of the following isincorrect?

Explanation

A problem like this can be solved similarly to a linear binomial like /

$-3 +8i \neq 3+8i$

Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Simplify the following expression, leaving no complex numbers in the denominator.

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

$\frac{1+5i}{8}$

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

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

$\frac{8+ 3i}{32}$

$\frac{20-4i}{15}$

Explanation

Solving this problem requires eliminating the nonreal term of the denominator. Our best bet for this is to cancel the nonreal term out by using the conjugate of the denominator.

Remember that for all binomials , there exists a conjugate such that $(a+b)\cdot(a-b) = (a^2-b^2)$ .

This can also be applied to complex conjugates, which will eliminate the nonreal portion entirely (since )!

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

$\frac{(3+2i)(4+4i)}{(4-4i)(4+4i)}$ Multiply both terms by the denominator's conjugate.

$\frac{(3+2i)(4+4i)}{(16-16i^2)} = \frac{(3+2i)(4+4i)}{32}$ Simplify. Note .

FOIL the numerator.

Combine and simplify.

$\frac{4+20i}{32} = \frac{1+5i}{8}$ Simplify the fraction.

Thus, $\frac{3+2i}{4-4i} = \frac{1+5i}{8}$ .

What is the sum of and given

and

Explanation

A complex number is a combination of a real and imaginary number. To add complex numbers, add each element separately.

In equation , is the real component and is the imaginary component (designated by ).

In equation , is the real component and is the imaginary component.

When added,

$a + b = \left ( 5+3i\right ) + \left ( 2 + i\right ) = \left ( 5 + 2\right ) + i\left ( 3 + 1\right ) = 7 + 4i$

What is the sum of and given

and

Explanation

A complex number is a combination of a real and imaginary number. To add complex numbers, add each element separately.

In equation , is the real component and is the imaginary component (designated by ).

In equation , is the real component and is the imaginary component.

When added,

$a + b = \left ( 5+3i\right ) + \left ( 2 + i\right ) = \left ( 5 + 2\right ) + i\left ( 3 + 1\right ) = 7 + 4i$

Complex numbers take the form , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Simplify the following expression, leaving no complex numbers in the denominator.

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

$\frac{1+5i}{8}$

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

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

$\frac{8+ 3i}{32}$

$\frac{20-4i}{15}$

Explanation

Solving this problem requires eliminating the nonreal term of the denominator. Our best bet for this is to cancel the nonreal term out by using the conjugate of the denominator.

Remember that for all binomials , there exists a conjugate such that $(a+b)\cdot(a-b) = (a^2-b^2)$ .

This can also be applied to complex conjugates, which will eliminate the nonreal portion entirely (since )!

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

$\frac{(3+2i)(4+4i)}{(4-4i)(4+4i)}$ Multiply both terms by the denominator's conjugate.

$\frac{(3+2i)(4+4i)}{(16-16i^2)} = \frac{(3+2i)(4+4i)}{32}$ Simplify. Note .

FOIL the numerator.

Combine and simplify.

$\frac{4+20i}{32} = \frac{1+5i}{8}$ Simplify the fraction.

Thus, $\frac{3+2i}{4-4i} = \frac{1+5i}{8}$ .

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