schoolAccedmy
Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.5

Solution

For z=x+iyz=x+iy:

r=∣z∣=x2+y2,θ=tan⁡−1(yx) (adjust quadrant)r=|z|=\sqrt{x^2+y^2},\quad \theta=\tan^{-1}\left(\frac{y}{x}\right)\ \text{(adjust quadrant)}

and

z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta)

(i) 4+3i4+3i

r=42+32=5,θ=tan⁡−1(34) r=\sqrt{4^2+3^2}=5,\quad \theta=\tan^{-1}\left(\frac{3}{4}\right) 4+3i=5(cos⁡(tan⁡−134)+isin⁡(tan⁡−134))\boxed{4+3i=5\left(\cos\left(\tan^{-1}\frac{3}{4}\right)+i\sin\left(\tan^{-1}\frac{3}{4}\right)\right)}

(ii) 1+i1+i

r=12+12=2,θ=π4 r=\sqrt{1^2+1^2}=\sqrt{2},\quad \theta=\frac{\pi}{4} 1+i=2(cos⁡π4+isin⁡π4)\boxed{1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)}

(iii) 12+32i\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i

r=(12)2+(32)2=1,θ=π3 r=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=1,\quad \theta=\frac{\pi}{3} 12+32i=cos⁡π3+isin⁡π3\boxed{\frac{1}{2}+\frac{\sqrt{3}}{2}i=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}

(iv) −52−532i-\dfrac{5}{2}-\dfrac{5\sqrt{3}}{2}i

r=(−52)2+(−532)2=5 r=\sqrt{\left(-\frac{5}{2}\right)^2+\left(-\frac{5\sqrt{3}}{2}\right)^2}=5

The point is in Quadrant III, and tan⁡−1(3)=π3\tan^{-1}(\sqrt{3})=\frac{\pi}{3}, so:

θ=π3+π=4π3\theta=\frac{\pi}{3}+\pi=\frac{4\pi}{3} −52−532i=5(cos⁡4π3+isin⁡4π3)\boxed{-\frac{5}{2}-\frac{5\sqrt{3}}{2}i=5\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)}

(v) 1−i1+i\dfrac{1-i}{1+i}

1−i1+i=1−i1+i⋅1−i1−i=(1−i)21−i2=1−2i+i22=−2i2=−i\begin{aligned} \frac{1-i}{1+i}&=\frac{1-i}{1+i}\cdot\frac{1-i}{1-i}=\frac{(1-i)^2}{1-i^2}\\ &=\frac{1-2i+i^2}{2}=\frac{-2i}{2}=-i \end{aligned}

So z=−iz=-i has r=1r=1 and θ=−π2\theta=-\frac{\pi}{2}.

−i=cos⁡(−π2)+isin⁡(−π2)\boxed{-i=\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)}

(vi) 3+i1+3i\dfrac{\sqrt{3}+i}{1+\sqrt{3}i}

3+i1+3i=(3+i)(1−3i)(1+3i)(1−3i)=23−2i4=32−12i\begin{aligned} \frac{\sqrt{3}+i}{1+\sqrt{3}i} &=\frac{(\sqrt{3}+i)(1-\sqrt{3}i)}{(1+\sqrt{3}i)(1-\sqrt{3}i)}\\ &=\frac{2\sqrt{3}-2i}{4}=\frac{\sqrt{3}}{2}-\frac{1}{2}i \end{aligned}

So r=1r=1 and θ=−π6\theta=-\frac{\pi}{6}.

3+i1+3i=cos⁡(−π6)+isin⁡(−π6)\boxed{\frac{\sqrt{3}+i}{1+\sqrt{3}i}=\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)}

(vii) 3+4i4+3i\dfrac{3+4i}{4+3i}

3+4i4+3i=(3+4i)(4−3i)(4+3i)(4−3i)=24+7i25=2425+725i\begin{aligned} \frac{3+4i}{4+3i} &=\frac{(3+4i)(4-3i)}{(4+3i)(4-3i)}\\ &=\frac{24+7i}{25}=\frac{24}{25}+\frac{7}{25}i \end{aligned} r=(2425)2+(725)2=1,θ=tan⁡−1(724) r=\sqrt{\left(\frac{24}{25}\right)^2+\left(\frac{7}{25}\right)^2}=1,\quad \theta=\tan^{-1}\left(\frac{7}{24}\right) 3+4i4+3i=cos⁡(tan⁡−1724)+isin⁡(tan⁡−1724)\boxed{\frac{3+4i}{4+3i}=\cos\left(\tan^{-1}\frac{7}{24}\right)+i\sin\left(\tan^{-1}\frac{7}{24}\right)}

(viii) 32+32i\dfrac{3}{2}+\dfrac{\sqrt{3}}{2}i

r=(32)2+(32)2=3,θ=π6 r=\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{3},\quad \theta=\frac{\pi}{6} 32+32i=3(cos⁡π6+isin⁡π6)\boxed{\frac{3}{2}+\frac{\sqrt{3}}{2}i=\sqrt{3}\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)}
arrow_back
Previous Question
Question 1
Next Question
Question 3
arrow_forward