Hi. Write the complex number √2 - i in the trigonometric form r(cosθ + i sinθ), with θ in the interval [0°, 360°). Thanks in advance!

Question

owlfred · Answer

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anonymous · Answer

root3(cos(315 + isin315)

anonymous · Answer

you need r and theta 
$$r=\sqrt{a^2+b^2}=\sqrt{\sqrt{2}^2+1}=\sqrt{3}$$\]

anonymous · Answer

to find theta  you know what 
$$tan(\theta)=\frac{b}{a}=\frac{-1}{\sqrt{2}}$$

anonymous · Answer

you are in quadrant IV  since you are to the right two and down one. so as himi1618 said $$\theta=315$$

anonymous · Answer

$$\sqrt{3}$$*(cos(315) + isin)315))
to find the quadrant , observe the sign in the given complex number. sqrt(2)-i, so its (+,-) and hence in 4th quadrant.

anonymous · Answer

and you are done
$$\sqrt{2}-i=\sqrt{3}(\cos(315)+isin(315))$$

anonymous · Answer

i have to add that it is somewhat objectionable to use degrees in these problems but if that is what it said, that is what you do

anonymous · Answer

OK thanks for your help. Just one question: should I just now use a calculator to find theta in degrees? These always confuse me...