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If (1+i)22-i = x + iy, then find the value of x + y.

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Question

If `(1 + i)^2/(2 - i)` = x + iy, then find the value of x + y.

Sum
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Solution

Given that: `(1 + i)^2/(2 - i)` = x + iy

⇒ `(1 + i^2 + 2i)/(2 - i)` = x + iy

⇒ `(1 - 1 + 2i)/(2 - i)` = x + iy

⇒ `(2i)/(2 - i)` = x + iy

⇒ `(2i(2 + i))/((2 - i)(2 + i))` = x + iy

⇒ `(4i + 2i^2)/(4 - i^2)` = x + iy

⇒ `(4i - 2)/(4 + 1)` = x + iy   ......[∵ i2 = –1]

⇒ `(-2 + 4i)/5` = x + iy

⇒ `(-2)/5 + 4/5 i` = x + iy

Comparing the real and imaginary parts,

We get x = `(-2)/5` and y = `4/5`

Hence, x + y = `(-2)/5 + 4/5 = 2/5`.

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Chapter 5: Complex Numbers and Quadratic Equations - Exercise [Page 91]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 5 Complex Numbers and Quadratic Equations
Exercise | Q 4 | Page 91

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