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Chapter 6: Problem 59

Suppose \(z\) is a complex number. Show that \(\frac{z-\bar{z}}{2 i}\) equals the imaginary part of \(z\).

Short Answer

Expert verified
The expression \(\frac{z-\bar{z}}{2 i}\) simplifies to \(b\), which is the imaginary part of the complex number \(z = a + bi\). Therefore, the given expression indeed equals the imaginary part of the complex number \(z\).

Step by step solution

01

Write z and its conjugate

Recall that a complex number \(z = a + bi\), where \(a\) and \(b\) are real numbers and \(i^2 = -1\). The conjugate of a complex number, denoted as \(\bar{z}\), is the complex number obtained by changing the sign of the imaginary part, i.e., \(\bar{z} = a - bi\).
02

Substitute z and its conjugate in the expression

Now, let's substitute the expressions for \(z\) and \(\bar{z}\) into the given expression \(\frac{z-\bar{z}}{2 i}\). This results in: \(\frac{(a + bi) - (a - bi)}{2 i}\)
03

Perform the subtraction in the numerator

We can simplify the numerator by subtracting the real and imaginary parts of the complex numbers separately. This gives: \(\frac{a + bi - a + bi}{2 i}\) From here, we can see that the real parts (\(a\)) cancel each other out, leaving: \(\frac{2 bi}{2 i}\)
04

Simplify the expression

Finally, we can further simplify the expression by canceling out the common factor of 2 and the imaginary unit \(i\): \(\frac{2 bi}{2 i} = b\)
05

Conclusion

We have shown that the expression \(\frac{z-\bar{z}}{2 i}\) simplifies to \(b\), which is the imaginary part of the complex number \(z = a + bi\). Therefore, the given expression indeed equals the imaginary part of the complex number \(z\).

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Most popular questions from this chapter

Explain why \(\left(\cos 1^{\circ}+i \sin 1^{\circ}\right)^{360}=1\).

Suppose \(z\) is a complex number. Show that \(\bar{z}=-z\) if and only if the real part of \(z\) equals 0 .

Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are vectors, neither of which is \(\mathbf{0}\). Show that \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}|\) if and only if \(\mathbf{u}\) and \(\mathbf{v}\) have the same direction.

Show that if \(z\) is a complex number, then the imaginary part of \(z\) is in the interval \([-|z|,|z|]\).

Using coordinates, show that if \(s\) and \(t\) are scalars and \(\mathbf{u}\) is a vector, then $$ (s t) \mathbf{u}=s(t \mathbf{u}) $$.
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