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Question: If a complex number z is such that \(z = 1 + i\), then the multiplicative inverse of \({z^2}\)is (wh...

If a complex number z is such that z=1+iz = 1 + i, then the multiplicative inverse of z2{z^2}is (where i=−1i = \sqrt { - 1} ).
(a) 2i (b) 1 - i (c) −i2 (d) i2  (a){\text{ 2i}} \\\ (b){\text{ 1 - i}} \\\ (c){\text{ }}\dfrac{{ - i}}{2} \\\ (d){\text{ }}\dfrac{i}{2} \\\

Explanation

Solution

In this question first of all compute z2{z^2} using the given complex number z. Now use the concept that multiplicative inverse of any number or of any complex number is simply the reciprocal of that number to obtain the multiplicative inverse.

Complete step-by-step answer:

Given function
z=1+iz = 1 + i
Now first find out z2.
So, z2=(1+i)2{z^2} = {\left( {1 + i} \right)^2}
So expand the square using the identity (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab
⇒z2=1+i2+2i\Rightarrow {z^2} = 1 + {i^2} + 2i
Now as we know that [i=−1⇒i2=−1]\left[ {i = \sqrt { - 1} \Rightarrow {i^2} = - 1} \right] so use this property in above equation we have,
⇒z2=1−1+2i=2i\Rightarrow {z^2} = 1 - 1 + 2i = 2i
Now as we know multiplicative inverse is the reciprocal of a number.
For example multiplicative inverse of 15 is (1/15).
So the multiplicative inverse (M.I) of z2 = 12i\dfrac{1}{{2i}}
⇒M.I=12i\Rightarrow M.I = \dfrac{1}{{2i}}
Now multiply and divide by (i) we have.
⇒M.I=12i×ii=i2i2=−i2\Rightarrow M.I = \dfrac{1}{{2i}} \times \dfrac{i}{i} = \dfrac{i}{{2{i^2}}} = \dfrac{{ - i}}{2}
So this is the required multiplicative inverse.
Hence option (C) is correct.

Note: There is a very standard procedure to find the multiplicative inverse of any complex number, we simply need to rationalize the reciprocal of complex number by multiplying both the numerator and the denominator by its complex conjugate, which is another complex number with the opposite sign between the real and the imaginary part of the original number.