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Chapter 2: Problem 138

The value of \((-i)^{(-1)}=\) (a) \(-(\pi / 2)\) (b) \((\pi / 2)\) (c) \(\mathrm{e}^{-(\pi / 2)}\) (b) \(\mathrm{e}^{(\pi / 2)}\)

Short Answer

Expert verified
\( \mathrm{e}^{-(\pi / 2)} \)

Step by step solution

01

Express -i in exponential form

The exponential form of a complex number is given by \(re^{i\theta}\), where r is the magnitude of the complex number and \(\theta\) is the argument (angle) of the complex number. Here, we have -i, which can be written as \(0 - i\). The magnitude of -i is: \[r = \sqrt{0^2 + (-1)^2} = 1\] Now we need to find the angle \(\theta\). Since -i is along the negative imaginary axis, the angle is \(\frac{3\pi}{2}\). So, the exponential form of -i is: \[-i = 1e^{i\frac{3\pi}{2}}\]
02

Raise the exponential form of -i to -1

Now we need to raise this to the power of -1: \[(-i)^{-1} = (1e^{i\frac{3\pi}{2}})^{-1}\] We know that \((a^b)^c = a^{bc}\), so we can multiply exponents: \[(1e^{i\frac{3\pi}{2}})^{-1} = 1e^{-i\frac{3\pi}{2}}\] Comparing this result to the given options, we can see that the answer is (c), which is \(\mathrm{e}^{-(\pi / 2)}\)

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Most popular questions from this chapter

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

Let \(z_{1}\) and \(z_{2}\) be two roots of equation \(z^{2}+a z+b=0 . Z\) is complex number. Assume that origin, \(\mathrm{z}_{1}\) and \(\mathrm{z}_{2}\) from an equilateral triangle then (a) \(\mathrm{a}^{2}=2 \mathrm{~b}\) (b) \(\mathrm{a}^{2}=3 \mathrm{~b}\) (c) \(a^{2}=4 b\) (b) \(a^{2}=b\)

If \(z=x+\) iy, \(x, y \in R\) and \(|x|+|y| \leq k|z|\) then \(k=\) (a) 1 (b) \(\sqrt{2}\) (c) \(\sqrt{3}\) (d) \(\sqrt{4}\)

If \(1, \mathrm{w}\) and \(\mathrm{w}^{2}\) are cube root of 1 then \((1-\mathrm{w})\left(1-\mathrm{w}^{2}\right)\) \(\left(1-\mathrm{w}^{4}\right)\left(1-\mathrm{w}^{8}\right)=\) (a) 16 (b) 8 (c) 9 (d) 64

If \(z=x+y\) i and \(|3 z|=|z-4|\) then \(x^{2}+y^{2}+x=\) (a) 1 (b) \(-1\) (c) 2 (d) \(-2\)
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