Find the multiplicative inverse of 3 + 7 i .

Let z = 3 + 7 i Since we know the multiplicative inverse of z is 1 z ⇒ z - 1 = 1 z = 1 3 + 7 i × 3 - 7 i 3 - 7 i (multiply and divide by conjugate of denominator) = ( 3 - 7 i ) 9 - 7 i 2 = ( 3 - 7 i ) 9 + 7 ∵ i 2 = - 1 = ( 3 - 7 i ) 16 Hence, the multiplicative inverse of 3 + 7 i is 3 16 - 7 16 i .

Find the multiplicative inverse of 3 + 7 i .

Let z = 3 + 7 i Since we know the multiplicative inverse of z is 1 z ⇒ z - 1 = 1 z = 1 3 + 7 i × 3 - 7 i 3 - 7 i (multiply and divide by conjugate of denominator) = ( 3 - 7 i ) 9 - 7 i 2 = ( 3 - 7 i ) 9 + 7 ∵ i 2 = - 1 = ( 3 - 7 i ) 16 Hence, the multiplicative inverse of 3 + 7 i is 3 16 - 7 16 i .

Express the complex number 5 + 9 i - 3 + 4 i , in the standard form a + i b .

Let z = 5 + 9 i - 3 + 4 i = 5 + 9 i - 3 + 4 i × ( - 3 - 4 i ) ( - 3 - 4 i ) (multiply and divide by conjugate of denominator) = - 15 - 20 i - 27 i - 36 i 2 9 - 16 i 2 = - 15 - 47 i + 36 9 + 16 ∵ i 2 = - 1 = 21 - 47 i 25 So, the standard form is 21 25 - 47 25 i .

Express the complex number ( - 1 + 3 i ) - 1 , in the standard form a + i b .

Let z = ( - 1 + 3 i ) - 1 ⇒ z = 1 ( - 1 + 3 i ) ∵ z - 1 = 1 z = 1 ( - 1 + 3 i ) × ( - 1 - 3 i ) ( - 1 - 3 i ) (multiply and divide by conjugate of denominator) = ( - 1 - 3 i ) ( 1 - 3 i 2 ) = ( - 1 - 3 i ) ( 1 + 3 ) ∵ i 2 = - 1 = ( - 1 - 3 i ) 4 So, the standard form is - 1 4 - 3 4 i .

Express the complex number ( 2 + 3 i ) 2 2 + i , in the standard form a + i b .

Let z = ( 2 + 3 i ) 2 2 + i = 4 + 9 i 2 + 12 i 2 + i = 4 - 9 + 12 i 2 + i = - 5 + 12 i 2 + i = - 5 + 12 i 2 + i × 2 - i 2 - i (multiply and divide by conjugate of denominator) = - 10 + 5 i + 24 i - 12 i 2 4 - i 2 = - 10 + 5 i + 24 i + 12 4 + 1 ∵ i 2 = - 1 = 2 + 29 i 5 So, the standard form is 2 5 + 29 5 i .

Express the complex number 1 ( 2 + 3 i ) ( 1 - 2 i ) , in the standard form a + i b .

Let z = 1 ( 2 + 3 i ) ( 1 - 2 i ) = 1 2 - 4 i + 3 i - 6 i 2 = 1 2 - 4 i + 3 i + 6 ∵ i 2 = - 1 = 1 8 - i = 1 8 - i × 8 + i 8 + i (multiply and divide by conjugate of denominator) = 8 + i 64 - i 2 = 8 + i 64 + 1 ∵ i 2 = - 1 = 8 + i 65 So, the standard form is 8 65 + 1 65 i .

Express the complex number 2 + i ( 1 + i ) ( 1 - 2 i ) , in the standard form a + i b .

Let z = 2 + i ( 1 + i ) ( 1 - 2 i ) = 2 + i 1 - 2 i + i - 2 i 2 = 2 + i 1 - 2 i + i + 2 ∵ i 2 = - 1 = 2 + i 3 - i = 2 + i 3 - i × 3 + i 3 + i (multiply and divide by conjugate of denominator) = 6 + 2 i + 3 i + i 2 9 - i 2 = 6 + 2 i + 3 i - 1 9 + 1 ∵ i 2 = - 1 = 5 + 5 i 10 So, the standard form is 1 2 + 1 2 i .

Express the complex number ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i ) , in the standard form a + i b .

Let z = ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i ) = 6 + 9 i - 4 i - 6 i 2 2 - i + 4 i - 2 i 2 = 6 + 9 i - 4 i + 6 2 - i + 4 i + 2 ∵ i 2 = - 1 = 12 + 5 i 4 + 3 i = 12 + 5 i 4 + 3 i × 4 - 3 i 4 - 3 i (multiply and divide by conjugate of denominator) = 48 - 36 i + 20 i - 15 i 2 16 - 9 i 2 = 48 - 36 i + 20 i + 15 16 + 9 ∵ i 2 = - 1 = 63 - 16 i 25 So, the standard form is 63 25 - 16 25 i .

Express the complex number 3 + - 1 2 - - 1 , in the standard form a + i b . Also find the conjugate and modulus of each number.

Let z = 3 + - 1 2 - - 1 = 3 + i 2 2 - i 2 ∵ i 2 = - 1 = 3 + i 2 - i = 3 + i 2 - i × 2 + i 2 + i (multiply and divide by conjugate of denominator) = 6 + 3 i + 2 i + i 2 4 - i 2 = 6 + 3 i + 2 i - 1 4 + 1 ∵ i 2 = - 1 = 5 + 5 i 5 = 1 + i . So, the standard form is 1 + i . Therefore, the conjugate of 1 + i is 1 - i . We know that, the modulus of any complex number x + i y is x 2 + y 2 . Hence, the modulus of a complex number 1 + i is 1 2 + 1 2 . ⇒ 1 + i = 2 .