QuestionFind each of the following in rectangular form (a) $\sin (1-i)$ (c) $\ln \left(-e^{(1+i)}\right)$ (b) $i^{3 / 4}$ (d) $\ln (i+\sqrt{3})$

Studdy Solution

STEP 1

What is this asking? We need to rewrite some complex expressions in rectangular form, which means as $a + bi$ , where $a$ and $b$ are real numbers. Watch out! Complex functions can have multiple values!
We need to be careful with those tricky logarithms and fractional powers.

STEP 2

1. Sine of a complex number
2. Fractional power of $i$
3. Logarithm of a complex exponential
4. Logarithm of a complex number

STEP 3

Alright, let's rewrite the sine function using Euler's formula!
Remember, $\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$ .
So, $\sin(1-i) = \frac{e^{i(1-i)} - e^{-i(1-i)}}{2i}$ .

STEP 4

Let's simplify those exponents! $i(1-i) = i - i^2 = i - (-1) = 1+i$ .
Similarly, $-i(1-i) = -i + i^2 = -i - 1 = -1-i$ .

STEP 5

Substitute these back into our sine expression: $\sin(1-i) = \frac{e^{1+i} - e^{-1-i}}{2i}$ .

STEP 6

Expand using the properties of exponents: $\frac{e^1 e^i - e^{-1} e^{-i}}{2i} = \frac{e e^i - \frac{1}{e} e^{-i}}{2i}$ .

STEP 7

Now, let's use Euler's formula again, but this time for $e^i$ and $e^{-i}$ ! $e^i = \cos(1) + i\sin(1)$ and $e^{-i} = \cos(-1) + i\sin(-1) = \cos(1) - i\sin(1)$ .

STEP 8

Substitute these back into our expression: $\frac{e(\cos(1) + i\sin(1)) - \frac{1}{e}(\cos(1) - i\sin(1))}{2i}$ .

STEP 9

Distribute and group terms: $\frac{(e - \frac{1}{e})\cos(1) + (e + \frac{1}{e})i\sin(1)}{2i}$ .

STEP 10

Divide both terms in the numerator by $2i$ : $\frac{(e - \frac{1}{e})\cos(1)}{2i} + \frac{(e + \frac{1}{e})i\sin(1)}{2i}$ .

STEP 11

Multiply the first term by $\frac{i}{i}$ (which is equal to 1) to get rid of $i$ in the denominator: $\frac{(e - \frac{1}{e})i\cos(1)}{-2} + \frac{(e + \frac{1}{e})\sin(1)}{2}$ .

STEP 12

Rearrange to match the rectangular form $a + bi$ : $\frac{(e + \frac{1}{e})\sin(1)}{2} - i\frac{(e - \frac{1}{e})\cos(1)}{2}$ .

STEP 13

We can rewrite $i$ using Euler's formula: $i = e^{i(\frac{\pi}{2} + 2\pi k)}$ , where $k$ is an integer.

STEP 14

Now, let's raise $i$ to the power of $\frac{3}{4}$ : $i^{3/4} = (e^{i(\frac{\pi}{2} + 2\pi k)})^{3/4} = e^{i(\frac{3\pi}{8} + \frac{3\pi k}{2})}$ .

STEP 15

Let's find the distinct values for $k = 0, 1, 2, 3$ : $e^{i\frac{3\pi}{8}}$ , $e^{i\frac{15\pi}{8}}$ , $e^{i\frac{27\pi}{8}}$ , $e^{i\frac{39\pi}{8}}$ .

STEP 16

Using Euler's formula, we can express these in rectangular form.
For example, for $k=0$ , we have $\cos(\frac{3\pi}{8}) + i\sin(\frac{3\pi}{8})$ .
We can do the same for the other values of $k$ .

STEP 17

Remember, $\ln(z) = \ln|z| + i\arg(z)$ .
Let's apply this to our expression: $\ln(-e^{1+i})$ .

STEP 18

First, let's rewrite $-e^{1+i}$ as $-1 \cdot e^{1+i}$ .
We know that $-1 = e^{i(\pi + 2\pi k)}$ , where $k$ is an integer.

STEP 19

So, $-e^{1+i} = e^{i(\pi + 2\pi k)} \cdot e^{1+i} = e^{1 + i(1+\pi + 2\pi k)}$ .

STEP 20

Now, we can take the logarithm: $\ln(-e^{1+i}) = 1 + i(1+\pi + 2\pi k)$ .

STEP 21

Let $z = i + \sqrt{3}$ .
We can rewrite $z$ in polar form: $z = re^{i\theta}$ , where $r = |z|$ and $\theta = \arg(z)$ .

STEP 22

We calculate $r = |i + \sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2$ .

STEP 23

We find $\theta = \arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$ .
So, $z = 2e^{i(\frac{\pi}{6} + 2\pi k)}$ .

STEP 24

Now, we can take the logarithm: $\ln(i+\sqrt{3}) = \ln(2e^{i(\frac{\pi}{6} + 2\pi k)}) = \ln(2) + i(\frac{\pi}{6} + 2\pi k)$ .

STEP 25

(a) $\frac{(e + \frac{1}{e})\sin(1)}{2} - i\frac{(e - \frac{1}{e})\cos(1)}{2}$ (b) $\cos(\frac{3\pi}{8} + \frac{3\pi k}{2}) + i\sin(\frac{3\pi}{8} + \frac{3\pi k}{2})$ for $k = 0, 1, 2, 3$ (c) $1 + i(1+\pi + 2\pi k)$ for integer $k$ (d) $\ln(2) + i(\frac{\pi}{6} + 2\pi k)$ for integer $k$

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