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NCERT Class XI Mathematics - Complex Numbers and Quadratic Equations - Solutions

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Question : 11 of 52
Marks: +1, -0
4 - 3i
Solution:  
Let z = 4 – 3i Then z‾\overline{z} = 4 + 3i ... (i)
and ∣z∣2|z|^2 = (4)2+(3)2(4)^2+(3)^2 = 25 ... (ii)
z−1z^{-1} = z‾∣z∣2\frac{\overline{z}}{|z|^2} = 4+3i25\frac{4+3i}{25} (By using (i) and (ii))
= 425+i325\frac{4}{25} + i\frac{3}{25}
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