What is the trigonometric form of $(- 1 + 6 i)$ $(-1+6i)$ ?

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Answer 1 · 2015-12-28T21:15:21

$\sqrt{37} ∠ - 1, 4$ $sqrt37/_-1,4$

A complex number $x + i y$ $x+iy$ in rectangular form may be converted to polar form $r ∠ θ$ $r/_theta$ (trigonometric form) as follows :

$r = \sqrt{x^{2} + y^{2}} and θ = {tan}^{- 1} (\frac{y}{x})$ $r=sqrt(x^2+y^2) and theta =tan^(-1)(y/x)$ .

So in this particular case we get :

$r = \sqrt{{(- 1)}^{2} + 6^{2}} = \sqrt{37}$ $r=sqrt((-1)^2+6^2)=sqrt37$

$θ = {tan}^{- 1} (\frac{6}{- 1}) = - 80, 5^{\circ} = - 1, 4 r a d$ $theta=tan^(-1)(6/(-1))=-80,5^@=-1,4rad$

What is the trigonometric form of (−1+6i) (-1+6i) ?