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How to find the reciprocal of a complex number?
Let z = x + iy be a non-zero complex number. Then
1z
= 1x+iy
= 1x+iy Γ xβiyxβiy, [Multiplying numerator and denominator by conjugate of denominator i.e., Multiply both numerator and denominator by conjugate of x + iy]
= xβiyx2βi2y2
= xβiyx2+y2
= xx2+y2 + i(βy)x2+y2
Clearly, 1z is equal to the multiplicative inverse of z. Also,
1z = xβiyx2+y2 = Β―z|z|2
Therefore, the multiplicative inverse of a non-zero complex z is equal to its reciprocal and is represent as
Re(z)|z|2 + i(βIm(z))|z|2= Β―z|z|2
Solved examples on reciprocal of a complex number:
1. If a complex number z = 2 + 3i, then find the reciprocal of z? Give your answer in a + ib form.
Solution:
Given z = 2 + 3i
Then, Β―z = 2 - 3i
And |z| = βx2+y2
= β22+(β3)2
= β4+9
= β13
Now, |z|2 = 13
Therefore, 1z = Β―z|z|2 = 2β3i13 = 213 + (-313)i, which is the required a + ib form.
2. Find the reciprocal of the complex number z = -1 + 2i. Give your answer in a + ib form.
Solution:
Given z = -1 + 2i
Then, Β―z = -1 - 2i
And |z| = βx2+y2
= β(β1)2+22
= β1+4
= β5
Now, |z|2= 5
Therefore, 1z = Β―z|z|2 = β1β2i5 = (-15) + (-25)i, which is the required a + ib form.
3. Find the reciprocal of the complex number z = i. Give your answer in a + ib form.
Solution:
Given z = i
Then, Β―z = -i
And |z| = βx2+y2
= β02+12
= β0+1
= β1
= 1
Now, |z|2= 1
Therefore, 1z = Β―z|z|2 = βi1 = -i = 0 + (-i), which is the required a + ib form.
Note: The reciprocal of i is its own conjugate - i.
11 and 12 Grade Math
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