How do you simplify #3(cos((7pi)/3)+isin((7pi)/3))div(cos(pi/2)+isin(pi/2))# and express the result in rectangular form?

Question

Liam Brennan · Accepted Answer

The answer is #=-3/2+i3sqrt3/2#

We need Euler's relation

#e^(itheta)=costheta+isintheta#

#cos(2/3pi)=-1/2#

#sin(2/3pi)=sqrt3/2#

Therefore,

#cos(7/6pi)+isin(7/6pi)=e^(7/6ipi)#

#cos(pi/2)+isin(pi/2)=e^(ipi/2)#

So,

#(3(cos(7/6pi)+isin(7/6pi)))/(cos(pi/2)+isin(pi/2))#

#=3e^(7/6ipi)/e^(ipi/2)#

#=3e^(ipi(7/6-1/2))#

#=3e^(2/3ipi)#

#=3cos(2/3pi)+i3sin(2/3pi)#

#=-3/2+i3sqrt3/2#

Xavier Alonso · Answer

undefined To simplify the expression $3(\cos\left(\frac{7\pi}{3}\right) + i\sin\left(\frac{7\pi}{3}\right)) \div (\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right))$ and express the result in rectangular form, we'll use Euler's formula:

$$ e^{i\theta} = \cos(\theta) + i\sin(\theta) $$

First, let's express both numerator and denominator in Euler's form:

Numerator:
$$ \cos\left(\frac{7\pi}{3}\right) + i\sin\left(\frac{7\pi}{3}\right) = e^{i\left(\frac{7\pi}{3}\right)} $$

Denominator:
$$ \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) = e^{i\left(\frac{\pi}{2}\right)} $$

Now, let's rewrite the expression using these forms:

$$ \frac{3e^{i\left(\frac{7\pi}{3}\right)}}{e^{i\left(\frac{\pi}{2}\right)}} $$

Using the properties of exponents, when dividing exponential terms with the same base, we subtract the exponents:

$$ 3e^{i\left(\frac{7\pi}{3} - \frac{\pi}{2}\right)} $$

$$ 3e^{i\left(\frac{14\pi}{6} - \frac{3\pi}{6}\right)} $$

$$ 3e^{i\left(\frac{11\pi}{6}\right)} $$

Now, we need to convert this expression back to rectangular form:

Using Euler's formula in reverse:
$$ e^{i\theta} = \cos(\theta) + i\sin(\theta) $$

We find:
$$ \cos\left(\frac{11\pi}{6}\right) + i\sin\left(\frac{11\pi}{6}\right) $$

$$ \frac{\sqrt{3}}{2} - \frac{1}{2}i $$

Thus, the simplified expression in rectangular form is $ \frac{3\sqrt{3}}{2} - \frac{3}{2}i $.