2023HSCQ7Which of the following statements about complex numbers is true? A. For all real numbers x,y,θx,y,\thetax,y,θ with x≠0x\ne0x, for some real number . B. For all non-zero complex numbers , if and , then . C. For all real with , and . D. For all real with and , .
2023TrialQ11 (a)By first writing 3+i\sqrt{3}+i3+i and 3−i\sqrt{3}-i3−i in exponential form, express (3+i)12+(3−i)12(\sqrt{3}+i)^{12}+(\sqrt{3}-i)^{12}(3+i)12+ in the form of aba^bab, where aaa and bbb are integers.2 marksShort answer
2023TrialQ12 (a) (iii)Let α\alphaα be a real number and suppose that zzz is a complex number such that z+1z=2cosα.z+\frac{1}{z}=2\cos\alpha.z+z1=2cosα. You may assume that zn+1zn=2cos(nα)z^n+\frac{1}{z^n}=2\cos(n\alpha)zn+zn1= for all positive integer nnn. Let ω=z+1z.\omega=z+\frac{1}{z}.ω=z+z1. Hence or otherwise, find all solutions of 16(cosα)4+8(cosα)3−12(cosα)2−4cosα+1=0.16(\cos\alpha)^4+8(\cos\alpha)^3-12(\cos\alpha)^2-4\cos\alpha+1=0.16(cosα)4+8(cosα)3−2 marksShort answer
2023HSCQ12 (d)Find the cube roots of 2−2i2-2i2−2i. Give your answer in exponential form.3 marksShort answer
2023TrialQ13 (a) (ii)By considering the roots of z3=42(1+i)z^3=4\sqrt{2}(1+i)z3=42(1+i), show that cos7π12+cosπ12=cosπ4.\cos\frac{7\pi}{12}+\cos\frac{\pi}{12}=\cos\frac{\pi}{4}.cos127π+cos12π2 marksShort answer
2022HSCQ11 (c) (i)Write the complex number −3+i-\sqrt3+i−3+i in exponential form.2 marksShort answer
2022TrialQ16 (a) (i)Using Euler’s theorem, show that if z=eiθ=cosθ+isinθz=e^{i\theta}=\cos\theta+i\sin\thetaz=eiθ=cosθ+isinθ, then zn+z−n=2cosnθ.z^n+z^{-n}=2\cos n\theta.zn+z−n=2cosnθ.1 markShort answer
2022TrialQ16 (a) (ii)Hence show that cos7θ=164(cos7θ+7cos5θ+21cos3θ+35cosθ).\cos^7\theta=\frac1{64}\left(\cos7\theta+7\cos5\theta+21\cos3\theta+35\cos\theta\right).cos7θ=641(cos7θ+7cos5θ+21cos3θ+35cosθ).2 marksShort answer
2022TrialQ16 (a) (i)Prove that cosθ+isinθ−1cosθ+isinθ+1=itanθ2.\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=i\tan\frac\theta2.cosθ+isinθ+1cosθ+isinθ−1=itan2θ.2 marksShort answer
2021HSCQ11 (a)The complex numbers z=2eiπ/2z=2e^{i\pi/2}z=2eiπ/2 and w=6eiπ/6w=6e^{i\pi/6}w=6eiπ/6 are given. Find zwzwzw, giving the answer in the form reiθre^{i\theta}reiθ.2 marksShort answer
2021HSCQ14 (c) (ii)By using part (i), or otherwise, show that Re(eiπ/10)=5+58.\operatorname{Re}\left(e^{i\pi/10}\right)=\sqrt{\frac{5+\sqrt5}{8}}.Re(eiπ/10)=85+5.3 marksShort answer
2020HSCQ13 (d) (ii)By expanding (eiθ+e−iθ)4(e^{i\theta}+e^{-i\theta})^4(eiθ+e−iθ)4, show that cos4θ=18(cos(4θ)+4cos(2θ)+3).\cos^4\theta=\frac18\bigl(\cos(4\theta)+4\cos(2\theta)+3\bigr).cos4θ=81(cos3 marksShort answer
2020HSCQ13 (d) (i)Show that for any integer nnn, einθ+e−inθ=2cos(nθ).e^{in\theta}+e^{-in\theta}=2\cos(n\theta).einθ+e−inθ=2cos(nθ).1 markShort answer