What is the trigonometric form of $- 8 - i$ $-8-i$ ?

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Answer 1 · 2018-02-23T10:20:17

$- (8 + i) \approx - \sqrt{58} (cos (0.12) + i sin (0.12))$ $-(8+i)~~-sqrt58(cos(0.12)+isin(0.12))$

$- 8 - i = - (8 + i)$ $-8-i=-(8+i)$

For a given complex number, $z = a + b i$ $z=a+bi$ , $z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$

$r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$

$θ = {tan}^{- 1} (\frac{b}{a})$ $theta=tan^-1(b/a)$

Let's deal with $8 + i$ $8+i$

$z = 8 + i = r (cos θ + i sin θ)$ $z=8+i=r(costheta+isintheta)$

$r = \sqrt{8^{2} + 1^{2}} = \sqrt{65}$ $r=sqrt(8^2+1^2)=sqrt65$

$θ = {tan}^{- 1} (\frac{1}{8}) \approx {0.12}^{c}$ $theta=tan^-1(1/8)~~0.12^c$

$- (8 + i) \approx - \sqrt{58} (cos (0.12) + i sin (0.12))$ $-(8+i)~~-sqrt58(cos(0.12)+isin(0.12))$

What is the trigonometric form of −8−i-8-i?