s1_95-100aSOLUTION

# s1_95-100aSOLUTION - ACM95a/100a Last updated October 7...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ACM95a/100a Last updated October 7, 2009 Problem Set I Solutions Prepared by Faisal Amlani 1. Show that, for a real number n , each of the two numbers z = 1 ± ni satisfies the equation z 2 − 2 z + 1 + n 2 = 0. SOLUTION : For z = 1 + i , z 2 − 2 z + 1 + n 2 = (1 + ni ) 2 − 2(1 + ni ) + 1 + n 2 = (1 + 2 ni − n 2 ) − 2 − 2 ni + 1 + n 2 = (1 + 2 ni − n 2 ) − (1 + 2 ni − n 2 ) = 0 . For z = 1 − i , z 2 − 2 z + 1 + n 2 = (1 − ni ) 2 − 2(1 − ni ) + 1 + n 2 = (1 − 2 ni − n 2 ) − 2 + 2 ni + 1 + n 2 = (1 − 2 ni − n 2 ) − (1 − 2 ni − n 2 ) = 0 . 2. Write the following complex numbers in the form a + ib (a) bracketleftbigg 2 + i 6 i − (1 − 2 i ) bracketrightbigg 2 (b) (1 − i ) 7 SOLUTION : (a) parenleftbigg 2 + i 6 i − (1 − i 2) parenrightbigg 2 = parenleftbigg 2 + i − 1 + 8 i parenrightbigg 2 = 3 + 4 i − 63 − 16 i = 3 + 4 i − 63 − 16 i parenleftbigg − 63 + 16 i − 63 + 16 i parenrightbigg = − 253 4225 − i 204 4225 . (b) (1 − i ) 7 = ((1 − i ) 2 ) 2 (1 − i ) 2 (1 − i ) = ( − 2 i ) 2 ( − 2 i )(1 − i ) = ( − 4)( − 2 − 2 i ) = 8 + 8 i....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

s1_95-100aSOLUTION - ACM95a/100a Last updated October 7...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online