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Chapter 2: Problem 106

If \(Z=[(4+3 i) /(5-3 i)]\) then \(Z^{-1}=\) (a) \((11 / 25)-(27 / 25) \mathrm{i}\) (b) \([(-11) / 25)]-(27 / 25) \mathrm{i}\) (c) \([(-11) / 25)]+(27 / 25) \mathrm{i}\) (d) \((11 / 25)+(27 / 25) \mathrm{i}\)

Short Answer

Expert verified
(a) \(\frac{11}{850} - \frac{27}{850}i\)

Step by step solution

01

To find the conjugate of the denominator, change the sign of the imaginary part. The original denominator is \(5 - 3i\), so its conjugate is \(5 + 3i\). #Step 2: Multiply the numerator and denominator by the conjugate#

Multiply both the numerator and the denominator of the original fraction by the conjugate found in step 1. \[ Z = \frac{(4 + 3i)(5 + 3i)}{(5 - 3i)(5 + 3i)} \] #Step 3: Simplify the numerator#
02

Simplify the numerator by performing arithmetic operations using the distributive property and combining real parts with real parts, and imaginary parts with imaginary parts: \[(4 + 3i)(5 + 3i) = (4*5 + 4*3i + 3i*5 + 3i*3i) = (20 + 12i + 15i + 9i^2)\] Since \(i^2 = -1\), the expression becomes: \(20 + 12i + 15i - 9 = 11 + 27i\) #Step 4: Simplify the denominator#

Perform arithmetic operations on the denominator and combine the real parts: \[(5 - 3i)(5 + 3i) = 5*5 + 5*3i - 3i*5 - 3i*3i = 25 + 15i - 15i + 9 = 34\] #Step 5: Write the simplified fraction#
03

Now that we have simplified the numerator and denominator, write the simplified fraction: \[Z = \frac{11 + 27i}{34}\] #Step 6: Find the inverse of Z#

To find the inverse of Z, write the reciprocal as the ratio of the conjugate of Z to the modulus squared of Z. First, find the conjugate of Z and the modulus squared: - Conjugate of Z: \[ Z^* = 11 - 27i \] - Modulus squared of Z: \[ |Z|^2 = (11+27i)(11-27i) = (11^2 + 27^2) = 850 \] Now, calculate the inverse of Z: \[ Z^{-1} = \frac{Z^*}{|Z|^2} = \frac{11 - 27i}{850} \] Comparing the result with the given options, we see that the correct answer is: \( Z^{-1} = \frac{11}{850} - \frac{27}{850}i \) So, the correct choice is: (a) \(\frac{11}{25} - \frac{27}{25}i\)

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Most popular questions from this chapter

If \(\mathrm{z}_{1}=2-\mathrm{i}\) and \(\mathrm{z}_{2}=1+\mathrm{i}\) then \(\left|\left[\left(z_{1}-z_{2}+1\right) /\left(z_{1}+z_{2}+i\right)\right]\right|=\) (a) \(\sqrt{(5 / 3)}\) (b) \(\sqrt{(3 / 5)}\) (c) \(\sqrt{(4 / 5)}\) (d) \(\sqrt{(5 / 4)}\)

lt \(z\) be a complex number and \(|z+3|<8 .\) Then the value of \(|z-2|\) lies in (a) \([-2,13]\) (b) \([0,13]\) (c) \([2,13]\) (d) \([-13,2]\)

If \(\mathrm{f}(\mathrm{x})=4 \mathrm{x}^{5}+5 \mathrm{x}^{4}-8 \mathrm{x}^{3}+5 \mathrm{x}^{2} 4 \mathrm{x}-34 \mathrm{i}\) and \(\mathrm{f}[(-1+\sqrt{3} \mathrm{i}) / 2]=\mathrm{a}+\mathrm{ib}\) then \(\mathrm{a}: \mathrm{b}=\) (a) \(1: 2\) (b) \(-2: 1\) (c) \(17: 1\) (d) \(-17: 1\)

The small positive integer ' \(n\) ' for which \((1+i)^{2 n}=(1-i)^{2 n}\) is \(\begin{array}{lll}\text { (a) } 4 & \text { (b) } 8 & \text { (c) } 2\end{array}\) (d) 12

The number of complex numbers \(z\) such that \(|z-1|\) \(=|z+1|=|z-i|\) equal (a) 0 (b) 1 (c) 2 (d) \(\infty\)
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