Question #161952

Given: 𝑧 = 6 + 7𝑖

Calculate the following:

1) 𝑧-4

2) 𝑧1/3, graph the roots.

3) ln 𝑧 , for 𝑛 = 1, 3, 5 & 7

4) 𝑓(𝑧) = 𝑧𝑧̅− 𝑧̅𝑖 + 5𝑖, and check your answer using 𝑧 = 𝑥 + 𝑖y


1
Expert's answer
2021-02-16T05:47:28-0500

1) z=6+7iz1=16+7i=67i85.z=6+7i\Rightarrow z^{-1}=\frac{1}{6+7i}=\frac{6-7i}{85}. Hence z4=(67i)4854=6827+i2184854.z^{-4}=\frac{(6-7i)^4}{85^4}=\frac{-6827+i2184}{85^4}.

2) z=85(685+i785).z=\sqrt{85}(\frac{6}{\sqrt{85}}+i\frac{7}{\sqrt{85}}). Hence, z=reiθz=re^{i\theta} where r=85,cos θ=685,sin θ=785.r=\sqrt{85}, cos \ \theta=\frac{6}{\sqrt{85}}, sin \ \theta =\frac{7}{\sqrt{85}}. Hence z1/3=r1/3eiθ/3,r1/3ei(θ/3+2π/3),r1/3ei(θ/3+4π/3).z^{1/3}=r^{1/3}e^{i\theta/3}, r^{1/3}e^{i(\theta/3+2\pi/3)}, r^{1/3}e^{i(\theta/3+4\pi/3)}. In other words, z1/3=ξ or ξω or ξω2z^{1/3}= \xi\ or\ \xi\omega \ or\ \xi \omega^2 where ξ=r1/3eiθ/3.\xi=r^{1/3}e^{i\theta/3} . (Diagram is attached below) Here ω\omega denotes complex cube root of unity.


3) ln(z)=lnz+iθ+i2πn=ln85+iθ+i2πn.ln (z)= ln |z|+i\theta+i2\pi n= ln \sqrt{85} +i\theta+i2\pi n. Now we put n=1,3,5,7.n=1,3,5,7. to get, ln(z)=ln85+iθ+i2π,ln85+iθ+i6πn,ln85+iθ+i10πn,ln85+iθ+i14πn.ln (z)= ln \sqrt{85} +i\theta+i2\pi , ln \sqrt{85} +i\theta+i6\pi n, ln \sqrt{85} +i\theta+i10\pi n, ln \sqrt{85} +i\theta+i14\pi n.4) f(z)=(6+7i)(67i)(67i)i+5i=78i.f(z)=(6+7i)(6-7i)-(6-7i)i+5i=78-i.

Verification: zz=(x+iy)(xiy)=x2+y2.z\overline{z}=(x+iy)(x-iy)=x^2+y^2. Also iz=ixy.i\overline{z}=ix-y. Hence, f(z)=x2+y2ixy+5i=x2+y2+y+i(5x).f(z)=x^2+y^2-ix-y+5i=x^2+y^2+y+i(5-x). Hence f(6+7i)=62+727+i(56)=f(6+7i)=6^2+7^2-7+i(5-6)= 78i.78-i.


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