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Chapter 2: Problem 101

If \(z=x+i y, x, y \in R\) and \(3 x+(3 x-y) i=4-6 i\) then \(z=\) (a) \((4 / 3)+\mathrm{i} 10\) (b) \((4 / 3)-\mathrm{i} 10\) (c) \([(-4) / 3]+\mathrm{i} 10\) (d) \([(-4) / 3]-\mathrm{i} 10\)

Short Answer

Expert verified
\( z = (\frac{4}{3})+ i 10 \)

Step by step solution

01

Equate Real and Imaginary Parts

Since we are given that \(3x+(3x-y)i=4-6i\), we can equate the real and imaginary parts. Real Parts: \(3x = 4\) Imaginary Parts: \(3x - y = -6\)
02

Solve for Real Part, x

To solve for the real part, \(x\), we have the following equation: \(3x = 4\) Divide both sides by 3: \( x = \frac{4}{3} \)
03

Solve for Imaginary Part, y

Now we know that \(x= \frac{4}{3}\). Let's use this value to find \(y\): \(3x - y = -6\) Substitute \(x = \frac{4}{3}\): \( 3 \left( \frac{4}{3} \right) - y = -6 \) Now, simplify the equation and isolate \(y\): \( 4 - y = -6 \) Add 6 to both sides: \(y = 4 + 6\) \(y = 10\)
04

Find the Correct Value for z

Now that we have the values for \(x\) and \(y\), we can write the complex number \(z\): \( z = \frac{4}{3} + 10i \) Now, let's compare this result with the given options: (a) \((\frac{4}{3})+ i 10\) (b) \((\frac{4}{3})- i 10\) (c) \([-(\frac{4}{3})]+i 10\) (d) \([-(frac{4}{3})]-i 10\) It matches option (a), therefore the correct solution is: \( z = (\frac{4}{3})+ i 10 \)

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Most popular questions from this chapter

Let \(z\) be complex number with modulus 2 and argument \([(-2 \pi) / 3]\) then \(\mathrm{z}=\) (a) \(-1+\mathrm{i} \sqrt{3}\) (b) \([(-1+i \sqrt{3}) / 2]\) (c) \(-1-\mathrm{i} \sqrt{3}\) (d) \([(-1-i \sqrt{3}) / 2]\)

The complex numbers \(\sin x+i \cos 2 x\) and \(\cos x-i \sin 2 x\) are conjugate of each other for (a) \(\mathrm{x}=\mathrm{k} \pi, \mathrm{k} \in \mathrm{z}\) (b) \(x=0\) (c) \(\mathrm{x}=[\mathrm{k}+(1 / 2)] \pi, \mathrm{k} \in \mathrm{z}\) (d) no value of \(\mathrm{x}\)

If the imaginary part of \([(2 z-3) /(i z+1)]\) is \(-2\) then the locus of the point representing \(z\) in the complex plane is (a) a circle (b) a straight line (c) a parabola (d) an ellipse

\((1+i)(2+a i)+(2+3 i)(3+i)=x+i y, x, x y \in R\) and \(x=y\) then \(\mathrm{a}=\) (a) 5 (b) \(-4\) (c) \(-5\) (d) 4

The principal argument of \(-2 \sqrt{3}-2 \mathrm{i}\) is (a) \([(-5 \pi) / 6]\) (b) \([(5 \pi) / 6]\) (c) \([(-2 \pi) / 3]\) (d) \([(2 \pi) / 3]\)
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