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Chapter 17: Problem 20

$$\frac{4+3 i}{3+4 i} \cdot \frac{3-4 i}{3-4 i}=\frac{24-7 i}{25}=\frac{24}{25}-\frac{7}{25} i$$

Short Answer

Expert verified
The simplified result is \( \frac{24}{25} - \frac{7}{25}i \).

Step by step solution

01

Understand the Exercise

The exercise requires simplifying the complex fraction \( \frac{4+3i}{3+4i} \) by multiplying it with \( \frac{3-4i}{3-4i} \), which is the conjugate of the denominator.
02

Multiply the Fractions

Multiply the given fraction by the conjugate. The product is \( \frac{(4+3i)(3-4i)}{(3+4i)(3-4i)} \). This uses the formula \((a+bi)(c-di) = (ac+bd) + (bc-ad)i\) for complex multiplication.
03

Simplify the Numerator

Calculate the numerator: \( (4+3i)(3-4i) = 4 \cdot 3 + 4 \cdot (-4i) + 3i \cdot 3 + 3i \cdot (-4i) = 12 - 16i + 9i - 12i^2 \). Since \( i^2 = -1 \), this becomes \( 12 - 16i + 9i + 12 = 24 - 7i \).
04

Simplify the Denominator

Calculate the denominator: \( (3+4i)(3-4i) = 3^2 - (4i)^2 = 9 - (-16) = 9 + 16 = 25 \). The denominator is a real number because it is the difference of squares.
05

Combine and Simplify

Combine the results from Step 3 and Step 4. The fraction is \( \frac{24-7i}{25} \). Divide each component separately: \( \frac{24}{25} - \frac{7}{25}i \).
06

Result Verification

Verify the result by comparing \( \frac{24-7i}{25} \) and \( \frac{24}{25} - \frac{7}{25}i \). They are equivalent, confirming the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Conjugate
A complex conjugate is a pair of complex numbers that have the same real part but an opposite imaginary part. If a complex number is written as \( a + bi \), its conjugate would be \( a - bi \). This concept is essential when simplifying complex fractions because it helps in making the denominator a real number.
  • Conjugates look similar but have a different sign for the imaginary component.
  • Multiplying a complex number by its conjugate eliminates the imaginary part in the denominator.
The reason this works is because the product of a complex number and its conjugate is always a real number. This product is given by the formula \((a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2\). Notice how the imaginary parts cancel each other out.
In the problem, multiplying \(3+4i\) by its conjugate \(3-4i\) changed the denominator to \(25\), a real number.
Complex Multiplication
Complex multiplication involves two complex numbers and follows similar basic principles to the multiplication of algebraic expressions. For two complex numbers \(x = a + bi\) and \(y = c + di\), their product is described using the distributive property:
  • Multiply the real parts: \(ac\).
  • Multiply the imaginary parts: \(bd\).
  • Combine the product of the real part of one and the imaginary part of the other to form the imaginary part: \((bc + ad)i\).
Thus, the multiplication results in \((a + bi)(c + di) = (ac - bd) + (ad + bc)i\).
In the exercise, multiplying \(4+3i\) by the conjugate \(3-4i\) yields \(24 - 7i\), using the formula described. This carefully combines both real and imaginary components uniquely to form another complex number.
Simplification of Complex Fractions
Simplifying a complex fraction requires manipulating both the numerator and the denominator to achieve a standard complex format. When simplifying, typically the complex number in the denominator is removed, turning it into a real number. This is achieved by multiplying by the conjugate:
  • The given complex fraction is multiplied by a fraction equivalent to 1, such as \(\frac{3-4i}{3-4i}\).
  • The multiplication is computed separately for both the numerator and the denominator.
  • The denominator becomes a real number, simplifying the entire fraction.
Finally, you separate the real and imaginary parts, expressing them as two distinct fractions/as two separate terms.
In the original exercise, the simplified form of the fraction \(\frac{4+3i}{3+4i}\) becomes \(\frac{24-7i}{25}\), further separated into \(\frac{24}{25} - \frac{7}{25}i\). This transformation simplifies calculations in subsequent mathematical operations.

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Most popular questions from this chapter

(a) If we take \(\arg \left(z_{1}\right)=\pi\) and \(\arg \left(z_{2}\right)=\pi / 2\) then \(\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=3 \pi / 2\) is an argument of the product \(z_{1} z_{2}=-5 i .\) With these same arguments we see that \(\arg \left(z_{1}\right)-\arg \left(z_{2}\right)=\pi / 2\) is an argument of the quotient \(z_{1} / z_{2}=\frac{1}{5} i\) (b) If we take \(\arg \left(z_{1}\right)=\pi\) and \(\arg \left(z_{2}\right)=-\pi / 2\) then \(\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=\pi / 2\) is an argument of the product \(z_{1} z_{2}=5 i .\) With these same arguments we see that \(\arg \left(z_{1}\right)-\arg \left(z_{2}\right)=3 \pi / 2\) is an argument of the quotient \(z_{1} / z_{2}=-\frac{1}{5} i\)

We have $$\lim _{\Delta z \rightarrow 0} \frac{\overline{z+\Delta z}-\bar{z}}{\Delta z}=\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}.$$ If we let \(\Delta z \rightarrow 0\) along a horizontal line then \(\Delta z=\Delta x, \overline{\Delta z}=\Delta x,\) and $$\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}=1.$$ If we let \(\Delta z \rightarrow 0\) along a vertical line then \(\Delta z=i \Delta y, \overline{\Delta z}=-i \Delta y,\) and $$\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}=\lim _{\Delta y \rightarrow 0} \frac{-i \Delta y}{i \Delta y}=-1.$$ Since these two limits are not equal, \(f(z)=\bar{z}\) cannot be differentiable at any \(z\).

\(u=x^{2}-y^{2}, \quad v=-2 x y ; \quad \frac{\partial u}{\partial x}=2 x, \quad \frac{\partial v}{\partial y}=-2 x ; \quad \frac{\partial u}{\partial y}=-2 y, \quad-\frac{\partial v}{\partial x}=2 y\) The Cauchy-Riemann equations hold only at \((0,0) .\) since there is no neighborhood about \(z=0\) within which \(f\) is differentiable we conclude \(f\) is nowhere analytic.

$$\cos \frac{12 \pi}{8}+i \sin \frac{12 \pi}{8}=-i$$

$$(2 \sqrt{2})^{5}\left[\cos \left(-\frac{5 \pi}{4}\right)+i \sin \left(-\frac{5 \pi}{4}\right)\right]=-128+128 i$$
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