Question #203452

Given that 𝑧1 = 3 + 𝑖 π‘Žπ‘›π‘‘ 𝑧2 = 2 βˆ’ 𝑖: i. Find the modulus and argument of 𝑧1/𝑧2 (5 marks) ii. Express 𝑧1/𝑧2 in polar and exponential form (2 marks) iii. Use de Moivre’s theorem to find an expression for ( 𝑧1/𝑧2) ^4 


1
Expert's answer
2021-06-07T17:20:31-0400

i)∣z1z2∣=∣3+i2βˆ’i∣=∣(3+i)(2+i)22βˆ’i2∣=∣5(1+i)5∣=∣1+i∣=12+12=2arg(z1z2)=arg(1+i)=tanβˆ’1(1)=Ο€4ii)Since,z=reiΞΈ.z1z2=1+i=2eiΟ€4.This is the exponential form.Polarform,z=reiΞΈ=r(cosΞΈ+isinΞΈ).z1z2=2eiΟ€4=2(cosΟ€4+isinΟ€4)iii)by using de Moivre’s theorem,(z1z2)4=(2)4ei4Ο€4=4eiΟ€=4(cosΟ€+isinΟ€)=4((βˆ’1)+i(0))=βˆ’4i)\\ |\frac{z_1}{z_2}|=|\frac{3+i}{2-i}|=|\frac{(3+i)(2+i)}{2^2-i^2}|=|\frac{5(1+i)}{5}|=|1+i|=\sqrt{1^2+1^2}=\sqrt2\\ arg(\frac{z_1}{z_2})=arg(1+i)=tan^{-1}(1)=\frac{\pi}{4}\\ ii)\\ Since, z=re^{i\theta}.\\ \frac{z_1}{z_2}=1+i=\sqrt2e^{i\frac{\pi}{4}}.\\ \text{This is the exponential form.}\\ Polar form, z=re^{i\theta}=r(cos\theta +isin\theta).\\ \frac{z_1}{z_2}=\sqrt2e^{i\frac{\pi}{4}}=\sqrt2({cos\frac{\pi}{4}}+isin{\frac{\pi}{4}})\\ iii)\\ \text{by using de Moivre's theorem,}\\ (\frac{z_1}{z_2})^4=(\sqrt2)^4e^{i4\frac{\pi}{4}} =4e^{i\pi} =4(cos\pi+isin\pi)=4((-1)+i(0))=-4


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