How do you convert $- 1 + i$ $-1+i$ to polar form?

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Answer 1 · 2016-05-22T12:42:09

Polar form of $- 1 + i$ $-1+i$ is $(\sqrt{2}, \frac{3 π}{4})$ $(sqrt2,(3pi)/4)$

A complex number $a + i b$ $a+ib$ in polar form is written as

$r cos θ + i r sin θ$ $rcostheta+irsintheta$ , $cos θ = \frac{a}{r}$ $costheta=a/r$ and $sin θ = \frac{b}{r}$ $sintheta=b/r$

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

As in $- 1 + i$ $-1+i$ $a = - 1$ $a=-1$ and $b = 1$ $b=1$

$r = \sqrt{{(- 1)}^{2} + 1^{2}} = \sqrt{2}$ $r=sqrt((-1)^2+1^2)=sqrt2$ and hence

$cos θ = - \frac{1}{\sqrt{2}}$ $costheta=-1/sqrt2$ and $sin θ = \frac{1}{\sqrt{2}}$ $sintheta=1/sqrt2$

Hence, $θ = \frac{3 π}{4}$ $theta=(3pi)/4$ and

Polar form of $- 1 + i$ $-1+i$ is $(\sqrt{2}, \frac{3 π}{4})$ $(sqrt2,(3pi)/4)$

How do you convert -1+i−1+i-1+i to polar form?