Theory of Intelligence
10TH•Mathematics•Complex Numbers•1.5–1.6 Real/Imaginary Parts and Simultaneous Equations

1.5 Finding Real and Imaginary Parts of Complex Number of $(x + iy)^n$ and $\left(\dfrac{x_1 + iy_1}{x_2 + iy_2}\right)^n$, where $x_2 + iy_2 \neq 0$ and $n = \pm 1$ and $n = \pm 2$

Real and imaginary parts of powers; simultaneous linear equations with complex coefficients; EXERCISE 1.4 and REVIEW EXERCISE 1.

1.5 Finding Real and Imaginary Parts of Complex Number of (x+iy)n(x + iy)^n and (x1+iy1x2+iy2)n\left(\dfrac{x_1 + iy_1}{x_2 + iy_2}\right)^n, where x2+iy2≠0x_2 + iy_2 \neq 0 and n=±1n = \pm 1 and n=±2n = \pm 2

(x+iy)n(x + iy)^n when n=±1n = \pm 1

Let z=(x+iy)nz = (x + iy)^n be a complex number.

(i) For n=1n = 1

We have z=(x+iy)1=x+iyz = (x + iy)^1 = x + iy

Re(z)=x,Im(z)=yRe(z) = x, Im(z) = y

(ii) For n=−1n = -1

We have z=(x+iy)−1z = (x + iy)^{-1}

=1x+iy=x−iy(x+iy)(x−iy)=x−iyx2+y2=xx2+y2−iyx2+y2= \dfrac{1}{x + iy} = \dfrac{x - iy}{(x + iy)(x - iy)} = \dfrac{x - iy}{x^2 + y^2} = \dfrac{x}{x^2 + y^2} - \dfrac{iy}{x^2 + y^2}

∴Re(z)=xx2+y2,Im(z)=−yx2+y2\therefore Re(z) = \dfrac{x}{x^2 + y^2}, Im(z) = \dfrac{-y}{x^2 + y^2}

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Skilled Practice!
Find real and imaginary parts of (x+iy)n(x + iy)^n for n=±2n = \pm 2

Example 17

Find real and imaginary parts of (4+3i)−1(4 + 3i)^{-1}.

Solution Let z=(4+3i)−1z = (4 + 3i)^{-1}

=14+3i=1(4−3i)(4+3i)(4−3i)=4−3i16+9=4−3i25=425−325i= \dfrac{1}{4 + 3i} = \dfrac{1(4 - 3i)}{(4 + 3i)(4 - 3i)} = \dfrac{4 - 3i}{16 + 9} = \dfrac{4 - 3i}{25} = \dfrac{4}{25} - \dfrac{3}{25}i

∴Re(z)=425,Im(z)=−325\therefore Re(z) = \dfrac{4}{25}, Im(z) = -\dfrac{3}{25}

Example 18

Find real and imaginary parts of z=(4+3i)−2z = (4 + 3i)^{-2}.

Solution z=(4+3i)−2z = (4 + 3i)^{-2}

=1(4+3i)2=1(4)2+(3i)2+2(4)(3i)=116−9+24i=17+24i= \dfrac{1}{(4 + 3i)^2} = \dfrac{1}{(4)^2 + (3i)^2 + 2(4)(3i)} = \dfrac{1}{16 - 9 + 24i} = \dfrac{1}{7 + 24i} =(7−24i)(7+24i)(7−24i)= \dfrac{(7 - 24i)}{(7 + 24i)(7 - 24i)}
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? Do you know?
Re(z)=z+z‾2Re(z) = \dfrac{z + \overline{z}}{2},
Im(z)=z−z‾2iIm(z) = \dfrac{z - \overline{z}}{2i}

=7−24i49+576=7−24i625=7625−24625i= \dfrac{7 - 24i}{49 + 576} = \dfrac{7 - 24i}{625} = \dfrac{7}{625} - \dfrac{24}{625}i

∴Re(z)=7625,Im(z)=−24625\therefore Re(z) = \dfrac{7}{625}, Im(z) = -\dfrac{24}{625}

Example 19

Find real and imaginary parts of z=(4+3i3+2i)−1z = \left(\dfrac{4 + 3i}{3 + 2i}\right)^{-1}.

Solution z=(4+3i3+2i)−1z = \left(\dfrac{4 + 3i}{3 + 2i}\right)^{-1}

=3+2i4+3i=(3+2i)(4−3i)(4+3i)(4−3i)=12−9i+8i−6i2(4)2+(3)2=12−i+616+9= \dfrac{3 + 2i}{4 + 3i} = \dfrac{(3 + 2i)(4 - 3i)}{(4 + 3i)(4 - 3i)} = \dfrac{12 - 9i + 8i - 6i^2}{(4)^2 + (3)^2} = \dfrac{12 - i + 6}{16 + 9} =18−i25=1825−125i= \dfrac{18 - i}{25} = \dfrac{18}{25} - \dfrac{1}{25}i

∴Re(z)=1825,Im(z)=−125\therefore Re(z) = \dfrac{18}{25}, Im(z) = -\dfrac{1}{25}

Example 20

Find real and imaginary parts of z=(1+i1−i)−2z = \left(\dfrac{1 + i}{1 - i}\right)^{-2}.

Solution z=(1+i1−i)−2z = \left(\dfrac{1 + i}{1 - i}\right)^{-2}

=(1−i1+i)2=(1−i1+i×1−i1−i)2=(1−i−i+i212−i2)2= \left(\dfrac{1 - i}{1 + i}\right)^2 = \left(\dfrac{1 - i}{1 + i} \times \dfrac{1 - i}{1 - i}\right)^2 = \left(\dfrac{1 - i - i + i^2}{1^2 - i^2}\right)^2 =(1−2i−11+1)2=(−2i2)2=(−i)2=−1= \left(\dfrac{1 - 2i - 1}{1 + 1}\right)^2 = \left(\dfrac{-2i}{2}\right)^2 = (-i)^2 = -1

∴Re(z)=−1,Im(z)=0\therefore Re(z) = -1, Im(z) = 0

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