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

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Answer 1 · 2018-04-10T18:42:14

$\sqrt{37} [cos (2.976) + i sin (2.976)]$ $color(blue)(sqrt(37)[cos(2.976)+isin(2.976)]$

The trigonometric for of a complex number $(a + b i)$ $( a+bi)$ is given by:

$z = r [cos (θ) + i sin (θ)]$ $z = r[cos(theta)+isin(theta)]$

Where:

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

$θ = arctan (\frac{b}{a})$ $theta=arctan(b/a)$

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

$θ = arctan (\frac{1}{- 6}) \approx - 0.1651486774$ $theta=arctan(1/-6)~~-0.1651486774$

This is in the IV quadrant.

Remember that $tan (θ)$ $tan(theta)$ only has an inverse for the domain:

$- \frac{π}{2} < θ < \frac{π}{2}$ $-pi/2 < theta < pi/2$

So $arctan (y)$ $arctan(y)$ will return angles in this range.

So we need to add $π$ $pi$ to this result, since $(- 6 + i)$ $(-6+i)$ is in the II quadrant:

$- 0.1651486774 + π = 2.976$ $-0.1651486774+pi=2.976$ 3 d.p.

So:

$θ = 2.976$ $theta=2.976$

Trigonometric form is therefore:

$\sqrt{37} [cos (2.976) + i sin (2.976)]$ $color(blue)(sqrt(37)[cos(2.976)+isin(2.976)]$

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