If Z = 7 + 3i and W = 1- i, find and simplify Z/W
Complex number is the sum of a real number and an imaginary number. These are the numbers that can be written in the form of a+ib, where a and b both are real numbers. It is denoted by z.
Here the value ‘a’ is called the real part which is denoted by Re(z), and ‘b’ is called the imaginary part Im(z). In complex numbers form a +bi, ‘i’ is an imaginary number called “iota”.
The value of i is (√-1) or we can write as i2 = -1.
For example:
3+11i is a complex number, where 3 is a real number (Re) and 11i is an imaginary number (Im).
2+12i is a complex number where 2 is a real number (Re) and 12i is an imaginary number (im)
The Combination of a real number and the imaginary number is called a Complex number.
Imaginary numbers
The numbers which are not real are termed Imaginary numbers. It gives a result in negative after squaring an imaginary number. Imaginary numbers are represented as Im().
Example: √-15, 9i, -21i are all imaginary numbers. Here ‘i’ is an imaginary number called “iota”
If Z = 7 + 3i and W = 1- i, find and simplify Z/W
Solution:
Given Z = 7 + 3i
W = 1- i.
To find Z/W, = (7 + 3i)/ (1- i)
simplifying it by multiplying the numerator and denominator with the conjugate of denominator
= {(7 + 3i )/ (1- i)} × {(1+i)/(1+i)}
= {(7+3i)(1+i)} / {(1-i)(1+i)}
= {7 +7i +3i + 3i2} / {1-(i)2}
= {7+10i -3} / {(1 +1)}
= (4+10i)/ 2
= 4/2 + 10/2i
= 2 + 5i
Similar Questions
Question 1: Let Z = 5 + 2i and W = 1- i. Find and simplify Z/W.
Solution:
Given Z = 5 + 2i
W = 1- i.
To find Z/W, = (5+ 2i)/ (1- i)
simplifying it by multiplying the numerator and denominator with the conjugate of denominator
= {(5 + 2i)/ (1- i) } × {(1+i)/(1+i)}
= {(5+2i)(1+i)} / {(1-i)(1+i)}
= {5 +5i +2i + 2i2} / {1-(i)2}
= {5+7i -2} / {(1 +1)}
= (3+7i)/ 2
= 3/2 + 7/2i
Question 2: Express in form of a+ib, 9(3+5i) + i(5+2i)
Solution:
Given: 9(3+5i) + i(5+2i)
= 27 +45i +5i +2i2
= 27 +50i + 2(-1)
= 27 +50i -2
= 25 + 50i
Question 3: Solve (2-4i) / (-5i)?
Solution:
Given : (2-4i) / (-5i)
here standard form of denominator is -5i = 0 – 5i
conjugate of denominator 0-5i = 0 +5i
Multiply with the conjugate
therefore, {(2-4i) / (0 -5i) } x { (0+5i )/( 0 +5i )}
= { (2-4i)(0+5i ) } / { 0 – (5i)2 }
= { 10i – 20i2 } / { 0 – (25(-1) ) }
= { 10i – 20 (-1) } / 25
= ( 10i + 20 ) / 25
= 20/ 25 + 10/25 i
= 4/5 + 2/5 i
Question 4: Perform the indicated operation and write the answer in standard form (5 + 9i)?
Solution:
Given : 1/5+9i
Multiplying with the conjugate of denominators. i.e 5+9i = 5-9i
= {1/(5+9i) } × {( 5-9i)/(5 -9i) }
= (5-9i ) / { (5)2 – (9i)2 }
= (5-9i)/ { 25 – 81(-1)}
= (5-9i) / (25+81)
= (5-9i) / 106
= 5/106 – 9/106 i
Question 5: Simplify ( -√3 + √-2 ) ( 2√3-i)
Solution:
Given : ( -√3 + √-2 ) ( 2√3-i)
= { (-√3 )( 2√3) – (-√3)(i) } + { (√-2 )( 2√3) – (√-2)(i)
= -6 + √3i + 2√6i – √2i2
= – 6 + (√3 + 2√6)i – √2i2
= -6 + (√3 + 2√6)i + √2
= (√2 – 6 ) + (√3 + 2√6)i
Question 6: Simplify: (3 – 4i)(5 – 5i).
Solution:
Given : (3 -4i)(5-5i)
= 15 -15i -20i +20i2
= 15 -15i -20i + 20(-1)
= 15 – 15i – 20i -20
= -5 – 35i
Question 7: Simplify (2 + 3i) / (7 + 2i)
Solution:
Multiplying with the conjugate of denominators.
= {(2 + 3i) x (7 – 2i)) / {(7 + 2i) x (7 – 2i)}
=(14 -4i +21i – 6i2 ) / {49 -(2i)2 }
=(14 -4i + 21i +6) / (49 +4)
=(20+ 17i) / 53
= 20/53 -17/53 i