If z 1 = 2 - i and z 2 = - 2 + i , find R e 1 z 1 z ¯ 1 .

Given, z 1 = 2 - i ⇒ z ¯ 1 = 2 + i and z 2 = - 2 + i Now, 1 z 1 z ¯ 1 = 1 2 - i 2 + i ⇒ 1 z 1 z ¯ 1 = 1 4 - i 2 ⇒ 1 z 1 z ¯ 1 = 1 4 + 1 ∵ i 2 = - 1 ⇒ 1 z 1 z ¯ 1 = 1 5 + i . 0 Therefore, this is of the form x + i y , x , y ∈ R and x and y are real and imaginary parts of a complex number. ∴ R e 1 z 1 z ¯ 1 = 1 5 .

Find the real values of x and y if x - 2 + ( y - 3 ) i 1 + i = 1 - 3 i .

Given, x - 2 + ( y - 3 ) i 1 + i = 1 - 3 i ⇒ x - 2 + ( y - 3 ) i = 1 - 3 i 1 + i ⇒ x - 2 + ( y - 3 ) i = 1 + i - 3 i - 3 i 2 ⇒ x - 2 + ( y - 3 ) i = 1 + i - 3 i + 3 ∵ i 2 = - 1 ⇒ x - 2 + ( y - 3 ) i = 4 - 2 i Now, equate real and imaginary parts on each side, we get ⇒ x - 2 = 4 and y - 3 = - 2 ⇒ x = 6 and y = 1 .

Find the real values of x and y if x - y i 2 + 3 i = x + 2 i 1 - i .

Given, x - y i 2 + 3 i = x + 2 i 1 - i ⇒ x - y i 2 + 3 i 1 - i = x + 2 i ⇒ 2 x + 3 x i - 2 y i - 3 y i 2 1 - i = x + 2 i ⇒ 2 x + 3 x i - 2 y i + 3 y 1 - i = x + 2 i ∵ i 2 = - 1 ⇒ 2 x - 2 x i + 3 x i - 3 x i 2 - 2 y i + 2 y i 2 + 3 y - 3 y i = x + 2 i ⇒ 2 x - 2 x i + 3 x i + 3 x - 2 y i - 2 y + 3 y - 3 y i = x + 2 i ∵ i 2 = - 1 ⇒ 5 x + y + x i - 5 y i = x + 2 i ⇒ 5 x + y + i x - 5 y = x + 2 i Now, equate real and imaginary parts on each side, we get ⇒ 5 x + y = x and x - 5 y = 2 ⇒ y = - 4 x and x - 5 y = 2 → 1 Put y = - 4 x in 1 , we get ⇒ x + 20 x = 2 ⇒ 21 x = 2 ⇒ x = 2 21 and y = - 8 21 .

If a + i b c + i d = p + i q , prove that a - i b c - i d = p - i q .

Given, a + i b c + i d = p + i q Now apply conjugate on each side ⇒ a + i b c + i d ¯ = p + i q ¯ ⇒ a + i b ¯ c + i d ¯ = p + i q ¯ ∵ z 1 z 2 ¯ = z 1 ¯ z 2 ¯ , z 2 ≠ 0 ⇒ a - i b c - i d = p - i q Hence, proved.

If a + i b c + i d = p + i q , prove that p 2 + q 2 = a 2 + b 2 c 2 + d 2 .

Given, a + i b c + i d = p + i q Now apply modulus on each side ⇒ a + i b c + i d = p + i q ⇒ a + i b c + i d = p + i q ∵ z 1 z 2 = z 1 z 2 , z 2 ≠ 0 ⇒ a 2 + b 2 c 2 + d 2 = p 2 + q 2 ∵ | x + i y | = x 2 + y 2 Squaring on both sides we get p 2 + q 2 = a 2 + b 2 c 2 + d 2 Hence, proved.