3.9 - Calculating with complex numbers

3.9 - Calculating with complex numbers

The equation x2+1=0x^2+1=0 has no real solution, but it becomes solvable when the number system is enlarged. You will calculate with the new number ii, separate real and imaginary parts, and use conjugate pairs to recover all the roots of a polynomial with real coefficients.

Real and imaginary parts

What would happen if a number could have square −1-1? Introduce ii, defined by i2=−1i^2=-1. Then ii and −i-i both solve x2+1=0x^2+1=0. The ordinary rules of expansion still apply; replace every occurrence of i2i^2 by −1-1.

Complex number

A complex number has Cartesian form z=a+ibz=a+ib, where a,b∈Ra,b\in\mathbb R and i2=−1i^2=-1. Its real part is Re⁡z=a\operatorname{Re}z=a, and its imaginary part is Im⁡z=b\operatorname{Im}z=b.

The imaginary part is a real number: for z=3−2iz=3-2i, it is −2-2, rather than −2i-2i. Real numbers are included by taking b=0b=0; a purely imaginary number has a=0a=0. In particular, 0=0+0i0=0+0i has both parts zero.

Equality of complex numbers

For real a,b,c,da,b,c,d,

a+ib=c+id⟺a=c and b=d.a+ib=c+id\quad\Longleftrightarrow\quad a=c\text{ and }b=d.

The definition of ii and this equality rule are not listed in MF19: learn them.

Why compare two parts? A complex number has a unique Cartesian representation. Equating just the real parts leaves the imaginary component unchecked.

Worked example 1: matching a negative imaginary part

Question. Real numbers a,ba,b satisfy (a+2)+i(2b−1)=5−3i(a+2)+i(2b-1)=5-3i. Find aa and bb.

Plan. Equate real parts and imaginary parts separately.

a+2=5real parts2b−1=−3imaginary partsa=3,b=−1.\begin{aligned} a+2&=5 &&\text{real parts}\\ 2b-1&=-3 &&\text{imaginary parts}\\ a&=3,\qquad b=-1. \end{aligned}

Answer. a=3a=3, b=−1b=-1.

Check. Substitution gives (3+2)+i[2(−1)−1]=5−3i(3+2)+i[2(-1)-1]=5-3i, as required.

Change only the sign of the imaginary part on the right. The real equation stays the same.

Worked example 2: matching a positive imaginary part

Question. Real numbers a,ba,b satisfy (a+2)+i(2b−1)=5+3i(a+2)+i(2b-1)=5+3i. Find aa and bb.

Plan. Use the same two comparisons, keeping the new sign.

a+2=5,2b−1=3⟹a=3,b=2.a+2=5,\qquad 2b-1=3 \quad\Longrightarrow\quad a=3,\quad b=2.

Answer. a=3a=3, b=2b=2.

Check. The left side is 5+i(4−1)=5+3i5+i(4-1)=5+3i. Only bb changed, agreeing with the unchanged real part.

Powers of ii can be simplified by grouping pairs: i3=i2i=−ii^3=i^2i=-i and i4=(i2)2=1i^4=(i^2)^2=1. These follow directly from the definition.

Watch out

The wrong line i2=1i^2=1 changes complex arithmetic completely. It is (−i)2=i2=−1(-i)^2=i^2=-1, while i(−i)=1i(-i)=1. Keep the sign outside a squared bracket under control.

Use both parts in this check.

Arithmetic and conjugates

Add or subtract matching parts: (3+2i)+(1−4i)=4−2i(3+2i)+(1-4i)=4-2i, while (3+2i)−(1−4i)=2+6i(3+2i)-(1-4i)=2+6i. For multiplication, expand the brackets and then use i2=−1i^2=-1.

For division, the useful partner of a+iba+ib is a−iba-ib:

(a+ib)(a−ib)=a2−i2b2=a2+b2.(a+ib)(a-ib)=a^2-i^2b^2=a^2+b^2.

The result is real, and is positive unless both aa and bb are zero.

Complex conjugate

The complex conjugate of z=a+ibz=a+ib is z∗=a−ibz^*=a-ib. Conjugation changes the sign of the imaginary part and leaves the real part unchanged. Thus z+z∗=2az+z^*=2a and zz∗=a2+b2zz^*=a^2+b^2.

Multiply numerator and denominator by the denominator's conjugate. This multiplies the fraction by 11, so its value stays the same.

Division in Cartesian form

For c+id≠0c+id\ne0,

a+ibc+id=(a+ib)(c−id)c2+d2.\frac{a+ib}{c+id}=\frac{(a+ib)(c-id)}{c^2+d^2}.

This conjugate method is not listed in MF19. Show the expansion and simplification, rather than just a calculator result.

Worked example 3: multiply and divide

Question. Let u=3+2iu=3+2i and v=1+iv=1+i. Find uvuv and u/vu/v in Cartesian form.

Plan. Expand the product; for the quotient, use v∗=1−iv^*=1-i.

uv=(3+2i)(1+i)=3+5i+2i2=1+5iuv=(3+2i)(1−i)(1+i)(1−i)=3−i−2i21−i2=5−i2.\begin{aligned} uv&=(3+2i)(1+i)=3+5i+2i^2=1+5i\\ \frac uv&=\frac{(3+2i)(1-i)}{(1+i)(1-i)}\\ &=\frac{3-i-2i^2}{1-i^2}=\frac{5-i}{2}. \end{aligned}

Answer. uv=1+5iuv=1+5i and u/v=5/2−i/2u/v=5/2-i/2.

Check. Multiplying the quotient by vv gives (5−i)(1+i)/2=(6+4i)/2=3+2i=u(5-i)(1+i)/2=(6+4i)/2=3+2i=u.

Now change only the sign of the imaginary part of vv. Its conjugate changes too.

Worked example 4: a conjugate denominator

Question. Let u=3+2iu=3+2i and v=1−iv=1-i. Find uvuv and u/vu/v in Cartesian form.

Plan. Expand as before, but multiply the quotient by (1+i)/(1+i)(1+i)/(1+i).

uv=(3+2i)(1−i)=3−i−2i2=5−iuv=(3+2i)(1+i)(1−i)(1+i)=1+5i2.\begin{aligned} uv&=(3+2i)(1-i)=3-i-2i^2=5-i\\ \frac uv&=\frac{(3+2i)(1+i)}{(1-i)(1+i)} =\frac{1+5i}{2}. \end{aligned}

Answer. uv=5−iuv=5-i and u/v=1/2+5i/2u/v=1/2+5i/2.

Check. (1+5i)(1−i)/2=(6+4i)/2=u(1+5i)(1-i)/2=(6+4i)/2=u. The real denominator is again 22.

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