Question #203412

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 iii. Use de Moivre’s theorem to find an expression for ( 𝑧1 𝑧2 ) 4 


1
Expert's answer
2021-06-07T13:24:29-0400

i.)z1=3+i,  z2=2βˆ’iz1z2=(3+i)(2βˆ’i)=6βˆ’3i+2i+1=7βˆ’i∣z1z2∣=72+(βˆ’1)2=50=52arg⁑(∣z1z2∣)=arctan⁑(βˆ’17)=3Ο€2+Ο€2βˆ’0.142=6.141ii.)z1z2=50e6.141iz1z2=50(cos⁑(6.141)+isin⁑(6.141))iii.(z1z2)4=2500(cos⁑(6.141Γ—4)+isin⁑(6.141Γ—4))=2500(cos⁑(25.56)+isin⁑(25.56))\displaystyle i.)\\ z_1 = 3 + i,\,\, z_2 = 2 - i\\ z_1 z_2 = (3 + i)(2 - i) = 6 - 3i + 2i + 1 = 7 - i\\ |z_1 z_2| = \sqrt{7^2 + (-1)^2} = \sqrt{50} = 5\sqrt{2}\\ \arg(|z_1 z_2|) = \arctan\left(\frac{-1}{7}\right) = \frac{3\pi}{2} + \frac{\pi}{2} - 0.142 = 6.141\\ ii.)\\ z_1 z_2 = \sqrt{50}e^{6.141i}\\ z_1 z_2 = \sqrt{50}(\cos(6.141) + i\sin(6.141))\\ iii. \\ (z_1 z_2)^4 = 2500(\cos(6.141\times 4) + i\sin(6.141\times 4)) = 2500(\cos(25.56) + i\sin(25.56))


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