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Chapter 6: Problem 8

Write $$ \frac{1}{7\left(\cos \frac{\pi}{9}+i \sin \frac{\pi}{9}\right)} $$ in polar form.

Short Answer

Expert verified
The given expression in polar form is \(\frac{1}{7}(\cos(\frac{\pi}{9}) - i\sin(\frac{\pi}{9}))\).

Step by step solution

01

Convert denominator into polar form

The denominator is a complex number given by \(7(\cos(\frac{\pi}{9}) + i\sin(\frac{\pi}{9}))\). This complex number is already in polar form: \(r (\cos(\theta) + i \sin(\theta))\) with \(r = 7\) and \(\theta = \frac{\pi}{9}\). So, the denominator is \(7(\cos(\frac{\pi}{9}) + i\sin(\frac{\pi}{9}))\).
02

Divide complex numbers

To divide the two complex numbers, we want to find the quotient of their moduli and subtract their arguments. The modulus of the fraction is \[ \frac{1}{7}. \] The argument of the fraction is \[ 0 - \frac{\pi}{9} = -\frac{\pi}{9}. \]
03

Write the expression in polar form

Now that we have the modulus and argument of the expression, we can write it in polar form: \[ \frac{1}{7}(\cos(-\frac{\pi}{9}) + i\sin(-\frac{\pi}{9})). \]
04

Use trigonometric identities to simplify the expression

To simplify the expression, we can use the identities \(\cos(-\theta) = \cos(\theta)\) and \(\sin(-\theta) = -\sin(\theta)\). Applying these identities to our expression, we get \[ \frac{1}{7}(\cos(\frac{\pi}{9}) - i\sin(\frac{\pi}{9})). \] So, the given expression in polar form is \[ \frac{1}{7}(\cos(\frac{\pi}{9}) - i\sin(\frac{\pi}{9})). \]

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Most popular questions from this chapter

Show that if \(\mathbf{u}\) and \(\mathbf{v}\) are vectors, then $$ 2\left(|\mathbf{u}|^{2}+|\mathbf{v}|^{2}\right)=|\mathbf{u}+\mathbf{v}|^{2}+|\mathbf{u}-\mathbf{v}|^{2} $$.

Show that \(\overline{w \cdot z}=\bar{w} \cdot \bar{z}\) for all complex numbers \(\mathcal{w}\) and \(z\)

Write each expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$ (5+6 i)(2+7 i) $$

Suppose $$ f(x)=a x^{2}+b x+c $$ where \(a \neq 0\) and \(b^{2}<4 a c\). Verify by direct substitution into the formula above that $$ f\left(\frac{-b+\sqrt{4 a c-b^{2}} i}{2 a}\right)=0 $$ and $$ f\left(\frac{-b-\sqrt{4 a c-b^{2}} i}{2 a}\right)=0 $$.

Show that addition of complex numbers is associative, meaning that $$ u+(w+z)=(u+w)+z $$ for all complex numbers \(u, w,\) and \(z\).
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