Question #349057

Write z1=3–√+i and z2=1−i in trigonometric form. Then the argument of the quotient z1/z2is given by

1
Expert's answer
2022-06-13T05:12:03-0400
r2=(3)2+(1)2=4r^2=(-\sqrt{3})^2+(1)^2=4r0=>r=4=2r\ge0=>r=\sqrt{4}=2

Quadrant II


tanθ=13=13\tan \theta=\dfrac{-1}{\sqrt{3}}=-\dfrac{1}{\sqrt{3}}θ=π+tan1(13)=5π6\theta=\pi+\tan^{-1}(-\dfrac{1}{\sqrt{3}})=\dfrac{5\pi}{6}z1=2(cos5π6+isin5π6)z_1=2(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6})





r2=(1)2+(1)2=2r^2=(1)^2+(-1)^2=2r0=>r=2r\ge0=>r=\sqrt{2}

Quadrant IV


tanθ=13=11=1\tan \theta=\dfrac{-1}{\sqrt{3}}=-\dfrac{1}{1}=-1θ=2π+tan1(1)=7π4\theta=2\pi+\tan^{-1}(-1)=\dfrac{7\pi}{4}z2=2(cos7π4+isin7π4)z_2=\sqrt{2}(\cos\dfrac{7\pi}{4}+i\sin\dfrac{7\pi}{4})

z1/z2z_1/z_2



θz1/z2=5π67π4=11π12\theta_{z_1/z_2}=\dfrac{5\pi}{6}-\dfrac{7\pi}{4}=-\dfrac{11\pi}{12}

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS