Porisma

Worked solution

Apply the nine modulus and argument rules

The appendix works with z=1+iz=1+i and, where two numbers are needed, z1=1+iz_1=1+i and z2=3+iz_2=\sqrt3+i. Follow its nine examples: (a) zz∗zz^*; (b) ∣z∗∣\lvert z^*\rvert; (c) arg⁡z∗\arg z^*; (d) ∣z1z2∣\lvert z_1z_2\rvert; (e) ∣z4∣\lvert z^4\rvert; (f) ∣z1/z2∣\lvert z_1/z_2\rvert; (g) arg⁡(z1z2)\arg(z_1z_2); (h) arg⁡(z4)\arg(z^4); (i) arg⁡(z1/z2)\arg(z_1/z_2).

CJC · Complex Numbers · Appendix C worked examples

(a) Product with the conjugate

Answer

22.

Apply the rule to the printed example

zz∗=∣1+i∣2=12+12=2zz^*=\lvert1+i\rvert^2=1^2+1^2=2

For real x,y, zz∗=∣z∣2=x2+y2zz^*=\lvert z\rvert^2=x^2+y^2. The source example uses z=1+i.

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(b) Modulus of a conjugate

Answer

2\sqrt2.

Apply the rule to the printed example

∣1−i∣=∣1+i∣=12+12=2\lvert1-i\rvert=\lvert1+i\rvert=\sqrt{1^2+1^2}=\sqrt2

Conjugation preserves distance from the origin.

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(c) Argument of a conjugate

Answer

−π/4-\pi/4.

Apply the rule to the printed example

arg⁡(1−i)=−arg⁡(1+i)=−tan⁡−1(1/1)=−π4\begin{aligned}&\arg(1-i)\\ & =-\arg(1+i)\\ & =-\tan^{-1}(1/1)\\ & =-\frac\pi4\end{aligned}

For this example the negated argument already lies in (−π,π](-\pi,\pi]. In general reduce the sign-reversed angle into that interval; on the negative real axis the principal argument is π for both z and its conjugate, not −π. Zero has no argument.

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(d) Modulus of a product

Answer

222\sqrt2.

Apply the rule to the printed example

∣(1+i)(3+i)∣=∣1+i∣∣3+i∣=(2)(2)=22\begin{aligned}&\lvert(1+i)(\sqrt3+i)\rvert\\ & =\lvert1+i\rvert\lvert\sqrt3+i\rvert\\ & =(\sqrt2)(2)\\ & =2\sqrt2\end{aligned}

The source’s second number is 3+i\sqrt3+i, not 3+i; its modulus is 2.

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(e) Modulus of an integer power

Answer

44.

Apply the rule to the printed example

∣(1+i)4∣=∣1+i∣4=(2)4=4\lvert(1+i)^4\rvert=\lvert1+i\rvert^4=(\sqrt2)^4=4

For positive integer n, ∣zn∣=∣z∣n\lvert z^n\rvert=\lvert z\rvert^n. Negative integer powers additionally require z≠0; unrestricted complex powers are not asserted by this rule.

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(f) Modulus of a quotient

Answer

2/2\sqrt2/2.

Apply the rule to the printed example

∣1+i3+i∣=∣1+i∣∣3+i∣=22\left\lvert\frac{1+i}{\sqrt3+i}\right\rvert=\frac{\lvert1+i\rvert}{\lvert\sqrt3+i\rvert}=\frac{\sqrt2}{2}

Quotient rules require a nonzero denominator. Here 3+i\sqrt3+i has modulus 2.

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(g) Argument of a product

Answer

5π/125\pi/12.

Apply the rule to the printed example

arg⁡[(1+i)(3+i)]=arg⁡(1+i)+arg⁡(3+i)=π4+π6=5π12\begin{aligned}&\arg[(1+i)(\sqrt3+i)]\\ & =\arg(1+i)+\arg(\sqrt3+i)\\ & =\frac\pi4+\frac\pi6\\ & =\frac{5\pi}{12}\end{aligned}

For nonzero factors, add arguments modulo 2π, then select the principal value. This example’s sum is already in the principal interval. The source row starts on 38 and its worked example continues on 39.

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(h) Argument of an integer power

Answer

π\pi.

Apply the rule to the printed example

arg⁡[(1+i)4]=4arg⁡(1+i)=4(π4)=π\begin{aligned}&\arg[(1+i)^4]\\ & =4\arg(1+i)\\ & =4\left(\frac\pi4\right)\\ & =\pi\end{aligned}

For nonzero z and integer n, multiply the argument then reduce modulo 2π. The source result is the included endpoint π because (1+i)^4=−4. Do not replace it by the excluded endpoint −π.

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(i) Argument of a quotient

Answer

π/12\pi/12.

Apply the rule to the printed example

arg⁡(1+i3+i)=arg⁡(1+i)−arg⁡(3+i)=π4−π6=π12\begin{aligned}&\arg\left(\frac{1+i}{\sqrt3+i}\right)\\ & =\arg(1+i)-\arg(\sqrt3+i)\\ & =\frac\pi4-\frac\pi6\\ & =\frac\pi{12}\end{aligned}

Both numbers must be nonzero for their arguments to exist. Subtract arguments modulo 2π, then select the principal value; this difference needs no adjustment.

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Solution notes

Principal arguments use (−π,π](-\pi,\pi]. Product, quotient and integer-power argument equalities are modulo 2π statements until reduced into that interval; the conjugate rule also needs the negative-real-axis endpoint qualification. Zero has no argument.