How do you write the trigonometric form of $- 3 - i$ $-3-i$ ?

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Answer 1 · 2017-08-01T13:44:06

The trigonometric form is $= \sqrt{10} ({cos 198.4}^{\circ} + i {sin 198.4}^{\circ})$ $=sqrt10(cos198.4^@+isin198.4^@)$

Let $z = - 3 - i$ $z=-3-i$

The trigonometris form is

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

If $z = a + i b$ $z=a+ib$

$z = | z | (\frac{a}{| z |} + \frac{b}{| z |} i)$ $z=|z|(a/|z|+b/|z|i)$

The modulus is

$| z | = \sqrt{a^{2} + b^{2}} = \sqrt{{(- 3)}^{2} + {(- 1)}^{2}} = \sqrt{10}$ $|z|=sqrt(a^2+b^2)=sqrt((-3)^2+(-1)^2)=sqrt10$

Therefore,

$z = \sqrt{10} (- \frac{3}{\sqrt{10}} - \frac{1}{\sqrt{10}} i)$ $z=sqrt10(-3/sqrt10-1/sqrt10i)$

$cos θ = - \frac{3}{\sqrt{10}}$ $costheta=-3/sqrt10$

and

$sin θ = - \frac{1}{\sqrt{10}}$ $sintheta=-1/sqrt10$

We are in the quadrant $I I I$ $III$

$θ = {198.4}^{\circ}$ $theta=198.4^@$

Therefore,

$z = \sqrt{10} ({cos 198.4}^{\circ} + i {sin 198.4}^{\circ}) = e^{i {198.4}^{\circ}}$ $z=sqrt10(cos198.4^@+isin198.4^@)=e^(i198.4^@)$

How do you write the trigonometric form of −3−i-3-i?