Porisma

Worked solution

Practise eight exact moduli and principal arguments

Determine, in exact form, the modulus and argument of:

(i) 1212; (ii) −3-3; (iii) 7 i\sqrt7\,i; (iv) −13i-13i; (v) 1+i1+i; (vi) 1−3 i1-\sqrt3\,i; (vii) −12+2i-\sqrt{12}+2i; (viii) −2−2 i-\sqrt2-\sqrt2\,i.

CJC · Self-Practice 6

(i)

Answer

Modulus 1212; principal argument 00.

122+02=12\sqrt{12^2+0^2}=12; positive real axis gives 0.

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(ii)

Answer

Modulus 33; principal argument π\pi.

(−3)2+02=3\sqrt{(-3)^2+0^2}=3; negative real axis gives π\pi.

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(iii)

Answer

Modulus 7\sqrt7; principal argument π/2\pi/2.

02+(7)2=7\sqrt{0^2+(\sqrt7)^2}=\sqrt7; positive imaginary axis gives π/2\pi/2.

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(iv)

Answer

Modulus 1313; principal argument −π/2-\pi/2.

02+(−13)2=13\sqrt{0^2+(-13)^2}=13; negative imaginary axis gives −π/2-\pi/2.

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(v)

Answer

Modulus 2\sqrt2; principal argument π/4\pi/4.

12+12=2\sqrt{1^2+1^2}=\sqrt2; quadrant I gives tan⁡−1(1)=π/4\tan^{-1}(1)=\pi/4.

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(vi)

Answer

Modulus 22; principal argument −π/3-\pi/3.

12+(−3)2=2\sqrt{1^2+(-\sqrt3)^2}=2; quadrant IV gives −tan⁡−1(3)=−π/3-\tan^{-1}(\sqrt3)=-\pi/3.

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(vii)

Answer

Modulus 44; principal argument 5π/65\pi/6.

(−12)2+22=4\sqrt{(-\sqrt{12})^2+2^2}=4. In quadrant II, α=tan⁡−1(2/12)=π/6\alpha=\tan^{-1}(2/\sqrt{12})=\pi/6, so θ=π−α=5π/6\theta=\pi-\alpha=5\pi/6.

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(viii)

Answer

Modulus 22; principal argument −3π/4-3\pi/4.

(−2)2+(−2)2=2\sqrt{(-\sqrt2)^2+(-\sqrt2)^2}=2. In quadrant III, α=π/4\alpha=\pi/4, so θ=−π+α=−3π/4\theta=-\pi+\alpha=-3\pi/4.

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Solution notes

Both copies print all eight modulus/argument pairs. The explanatory calculations here are authored from the printed values.