1.5 Finding Real and Imaginary Parts of Complex Number of (x+iy)n and (x2+iy2x1+iy1)n, where x2+iy2=0 and n=±1 and n=±2
(x+iy)n when n=±1
Let z=(x+iy)n be a complex number.
(i) For n=1
We have z=(x+iy)1=x+iy
Re(z)=x,Im(z)=y
(ii) For n=−1
We have z=(x+iy)−1
=x+iy1=(x+iy)(x−iy)x−iy=x2+y2x−iy=x2+y2x−x2+y2iy
∴Re(z)=x2+y2x,Im(z)=x2+y2−y
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Skilled Practice!
Find real and imaginary parts of (x+iy)n for n=±2
Example 17
Find real and imaginary parts of (4+3i)−1.
Solution Let z=(4+3i)−1
=4+3i1=(4+3i)(4−3i)1(4−3i)=16+94−3i=254−3i=254−253i
∴Re(z)=254,Im(z)=−253
Example 18
Find real and imaginary parts of z=(4+3i)−2.
Solution z=(4+3i)−2
=(4+3i)21=(4)2+(3i)2+2(4)(3i)1=16−9+24i1=7+24i1
=(7+24i)(7−24i)(7−24i)
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? Do you know?
Re(z)=2z+z,
Im(z)=2iz−z
=49+5767−24i=6257−24i=6257−62524i
∴Re(z)=6257,Im(z)=−62524
Example 19
Find real and imaginary parts of z=(3+2i4+3i)−1.
Solution z=(3+2i4+3i)−1
=4+3i3+2i=(4+3i)(4−3i)(3+2i)(4−3i)=(4)2+(3)212−9i+8i−6i2=16+912−i+6
=2518−i=2518−251i
∴Re(z)=2518,Im(z)=−251
Example 20
Find real and imaginary parts of z=(1−i1+i)−2.
Solution z=(1−i1+i)−2
=(1+i1−i)2=(1+i1−i×1−i1−i)2=(12−i21−i−i+i2)2
=(1+11−2i−1)2=(2−2i)2=(−i)2=−1
∴Re(z)=−1,Im(z)=0