How do you convert $- 6 + 6 i$ $-6+6i$ to polar form?

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Answer 1 · 2016-09-30T08:09:29

$- 6 + 6 i = 6 \sqrt{2} (cos (\frac{3}{4} π) + i sin (\frac{3}{4} π))$ $-6+6i=6sqrt 2(cos(3/4pi) +i sin (3/4pi))$

The general $θ \in Q_{2}$ $theta inQ_2$ is $2kpi+3/4pi, k=0, +-1, |=2, +-3, ...$

The conversion formula for z=x+iy=(x, y) is

$(x, y)=r(cos theta, sin theta)$ , giving

$r = sqrt(x^2+y^2)$ (the principal square root) $>=0$ ,

$cos theta =x/r and sin theta = y/r$

Here, $x = -6 and y = 6$ . So,

$rsqrt((-6)^2+6^2)=6sqrt 2$

$cos theta =-6/(6sqrt 2)=-1/sqrt 2 and sin theta=6/(6sqrt 2)=1/sqrt 2$ .

Note that there is no common principal value for both.

In $(0, 2pi)$ , the value is $3/4piin Q_2$ ., wherein

sine is positive and cosine is negative..

And so,

$-6+6i=6sqrt 2(cos(3/4pi) +i sin (3/4pi))$

The general $theta inQ_2$ is $2kpi+3/4pi, k=0, +-1, |=2, +-3, ...$

How do you convert -6+6i−6+6i-6+6i to polar form?