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Chapter 1: Problem 32

Find the complex conjugate of each of the following: (a) \(6+4 i, 7-5 i, 4+i,-3-i\) (b) \(6,-3,4 i,-9 i\)

Short Answer

Expert verified
The complex conjugates of the given complex numbers are: (a) \(6-4i, 7+5i, 4-i, -3+i\) (b) \(6, -3, -4i, 9i\)

Step by step solution

01

Finding the complex conjugate of 6+4i

To find the complex conjugate of \(6+4i\), you simply switch the sign of the imaginary part. In this case, change the positive 4i into a negative 4i: \((6+4i)^* = 6-4i\).
02

Finding the complex conjugate of 7-5i

To find the complex conjugate of \(7-5i\), switch the sign of the imaginary part. In this case, change the negative 5i into a positive 5i: \((7-5i)^* = 7+5i\).
03

Finding the complex conjugate of 4+i

To find the complex conjugate of \(4+i\), switch the sign of the imaginary part. In this case, change the positive i into a negative i: \((4+i)^* = 4-i\).
04

Finding the complex conjugate of -3-i

To find the complex conjugate of \(-3-i\), switch the sign of the imaginary part. In this case, change the negative i into a positive i: \((-3-i)^* = -3+i\).
05

Finding the complex conjugate of 6

Since 6 is a real number, its complex conjugate is itself because there is no imaginary part to change sign: \((6)^* = 6\).
06

Finding the complex conjugate of -3

Since -3 is a real number, its complex conjugate is itself because there is no imaginary part to change sign: \((-3)^* = -3\).
07

Finding the complex conjugate of 4i

To find the complex conjugate of \(4i\), switch the sign of the imaginary part. In this case, change the positive 4i into a negative 4i: \((4i)^* = -4i\).
08

Finding the complex conjugate of -9i

To find the complex conjugate of \(-9i\), switch the sign of the imaginary part. In this case, change the negative 9i into a positive 9i: \((-9i)^* = 9i\). In summary, the complex conjugates are: (a) \(6-4i, 7+5i, 4-i, -3+i\) (b) \(6, -3, -4i, 9i\)

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Most popular questions from this chapter

Let \(V\) denote the set of ordered pairs of real numbers. If $\left(a_{1}, a_{2}\right)\( and \)\left(b_{1}, b_{2}\right)\( are elements of \)\mathrm{V}$ and \(c \in R\), define $$ \left(a_{1}, a_{2}\right)+\left(b_{1}, b_{2}\right)=\left(a_{1}+b_{1}, a_{2} b_{2}\right) \quad \text { and } \quad c\left(a_{1}, a_{2}\right)=\left(c a_{1}, a_{2}\right) . $$ Is \(\mathrm{V}\) a vector space over \(R\) with these operations? Justify your answer.

Let $\mathrm{V}=\left\\{\left(a_{1}, a_{2}\right): a_{1}, a_{2} \in F\right\\}\(, where \)F\( is a field. Define addition of elements of \)\mathrm{V}$ coordinatewise, and for \(c \in F\) and $\left(a_{1}, a_{2}\right) \in \mathrm{V}$, define $$ c\left(a_{1}, a_{2}\right)=\left(a_{1}, 0\right) . $$ Is \(\mathrm{V}\) a vector space over \(F\) with these operations? Justify your answer.

Find a unit vector \(u\) orthogonal to: (a) \(v=[1,2,3]\) and \(w=[1,-1,2]\) (b) \(v=3 \mathbf{i}-\mathbf{j}+2 \mathbf{k}\) and \(w=4 \mathbf{i}-2 \mathbf{j}-\mathbf{k}\)

Show that for complex numbers \(z\) and \(w\) (a) \(\operatorname{Re} z=\frac{1}{2}(z+\bar{z})\) (b) \(\operatorname{Im} z=\frac{1}{2}(z-\bar{z})\) (c) \(z w=0\) implies \(z=0\) or \(w=0\)

Let \(u=(1,-2,4), v=(3,5,1), w=(2,1,-3) .\) Find: (a) \(3 u-2 v\) (b) \(5 u+3 v-4 w\) \(\begin{array}{llll}\text { (c) } u \cdot v, & u \cdot w, & v \cdot w ;\end{array}\) (d) \(\|u\|,\|v\|\) (e) \(\cos \theta,\) where \(\theta\) is the angle between \(u\) and \(v\) \((\mathrm{f}) \quad d(u, v)\) \((g) \quad \operatorname{proj}(u, v)\)
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