Question #262541

Given two complex numbers z1=1+i and z2= √(3)-1

a) Write z1/z2 in algebraic and polar forms.

b) Deduce the exact values of Cos 5pie/12 and Sin 5pie/12

c) What is the lowest positive value of integer n such that (z1/z2) is real


1
Expert's answer
2021-11-08T17:39:55-0500

a)


z1z2=1+i3i=(1+i)(3+i)31\dfrac{z_1}{z_2}=\dfrac{1+i}{\sqrt{3}-i}=\dfrac{(1+i)(\sqrt{3}+i)}{3-1}

=3+i+i314=314+3+14i=\dfrac{\sqrt{3}+i+i\sqrt{3}-1}{4}=\dfrac{\sqrt{3}-1}{4}+\dfrac{\sqrt{3}+1}{4}i

z1=1+i,z=12+12=2z_1=1+i, |z|=\sqrt{1^2+1^2}=\sqrt{2}

tanθ1=11=1\tan\theta_1=\dfrac{1}{1}=1

z1=1+i=2(cos(π/4)+isin(π/4))z_1=1+i=\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))

z2=3i,z2=(3)2+(1)2=2z_2=\sqrt{3}-i, |z_2|=\sqrt{(\sqrt{3})^2+(-1)^2}=2

tanθ2=13,θ2=π/6\tan\theta_2=\dfrac{-1}{\sqrt{3}}, \theta_2=-\pi/6

z2=3i=2(cos(π/6)+isin(π/6))z_2=\sqrt{3}-i=2(\cos(-\pi/6)+i\sin(-\pi/6))

z1z2=1+i3i=2(cos(π/4)+isin(π/4))2(cos(π/6)+isin(π/6))\dfrac{z_1}{z_2}=\dfrac{1+i}{\sqrt{3}-i}=\dfrac{\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))}{2(\cos(-\pi/6)+i\sin(-\pi/6))}

=22(cos(π/4+π/6)+isin(π/4+π/6))=\dfrac{\sqrt{2}}{2}(\cos(\pi/4+\pi/6)+i\sin(\pi/4+\pi/6))

=22(cos(5π/12)+isin(5π/12))=\dfrac{\sqrt{2}}{2}(\cos(5\pi/12)+i\sin(5\pi/12))


b)


314=22cos(5π/12)\dfrac{\sqrt{3}-1}{4}=\dfrac{\sqrt{2}}{2}\cos(5\pi/12)

cos(5π/12)=624\cos(5\pi/12)=\dfrac{\sqrt{6}-\sqrt{2}}{4}


3+14=22sin(5π/12)\dfrac{\sqrt{3}+1}{4}=\dfrac{\sqrt{2}}{2}\sin(5\pi/12)

sin(5π/12)=6+24\sin(5\pi/12)=\dfrac{\sqrt{6}+\sqrt{2}}{4}

c)


(z1z2)n=(22)n(cos(5πn12)+isin(5πn12))(\dfrac{z_1}{z_2})^n=(\dfrac{\sqrt{2}}{2})^n(\cos(\dfrac{5\pi n}{12})+i\sin(\dfrac{5\pi n}{12}))

If (z1z2)n(\dfrac{z_1}{z_2})^n is real, then sin(5πn2)=0.\sin(\dfrac{5\pi n}{2})=0.

5πn12=πm,mZ\dfrac{5\pi n}{12}=\pi m, m\in \Z

n=12m5n=\dfrac{12m}{5}

The lowest positive value of integer nn is n=12.n=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