Solve the equation z4=3−iz^4 = \sqrt{3} - i, giving the roots in the form reiθre^{i\theta} where r>0r > 0 and −π<θ≤π-\pi < \theta \le \pi. Show the roots on an Argand diagram and state the cartesian equation of a geometrical shape that the roots lie on.

Work through it with AI →

It won't just hand you the answer — it asks where you're stuck.How the checker works

Worked solution

4(i)

1. Express the complex number $\sqrt{3} - i$ in exponential form.

Let w=3−iw = \sqrt{3} - i.
Modulus r=(3)2+(−1)2=3+1=2r = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2
Argument θ=tan⁡−1(−13)=−π6\theta = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}
So, w=2e−iπ6w = 2e^{-i\frac{\pi}{6}}.

To find all roots, introduce the general argument by adding multiples of 2π2\pi:
z4=2ei(−π6+2kπ)z^4 = 2e^{i\left(-\frac{\pi}{6} + 2k\pi\right)} for k∈Zk \in \mathbb{Z}.

2. Take the 4th root to find the general expression for $z$.

z=[2ei(−π6+2kπ)]14z = \left[2e^{i\left(-\frac{\pi}{6} + 2k\pi\right)}\right]^{\frac{1}{4}}
z=214ei(−π24+kπ2)z = 2^{\frac{1}{4}} e^{i\left(-\frac{\pi}{24} + \frac{k\pi}{2}\right)}

3. Substitute integer values of $k$ to find the four distinct roots within the principal argument range $-\pi < \theta \le \pi$.

For k=0k = 0:
z1=214e−iπ24z_1 = 2^{\frac{1}{4}} e^{-i\frac{\pi}{24}}

For k=1k = 1:
z2=214ei(−π24+12π24)=214ei11π24z_2 = 2^{\frac{1}{4}} e^{i\left(-\frac{\pi}{24} + \frac{12\pi}{24}\right)} = 2^{\frac{1}{4}} e^{i\frac{11\pi}{24}}

For k=2k = 2:
z3=214ei(11π24+12π24)=214ei23π24z_3 = 2^{\frac{1}{4}} e^{i\left(\frac{11\pi}{24} + \frac{12\pi}{24}\right)} = 2^{\frac{1}{4}} e^{i\frac{23\pi}{24}}

For k=−1k = -1:
z4=214ei(−π24−12π24)=214e−i13π24z_4 = 2^{\frac{1}{4}} e^{i\left(-\frac{\pi}{24} - \frac{12\pi}{24}\right)} = 2^{\frac{1}{4}} e^{-i\frac{13\pi}{24}}

All four roots have arguments within the interval (−π,π](-\pi, \pi].

Answer

z=214e−iπ24, 214ei11π24, 214ei23π24, 214e−i13π24z = 2^{\frac{1}{4}} e^{-i\frac{\pi}{24}},\ 2^{\frac{1}{4}} e^{i\frac{11\pi}{24}},\ 2^{\frac{1}{4}} e^{i\frac{23\pi}{24}},\ 2^{\frac{1}{4}} e^{-i\frac{13\pi}{24}}

4(ii)

1. Show the roots on an Argand diagram.

From part (i), the roots are of the form z=214eiθz = 2^{\frac{1}{4}} e^{i\theta} where θ=−π24,11π24,23π24,−13π24\theta = -\frac{\pi}{24}, \frac{11\pi}{24}, \frac{23\pi}{24}, -\frac{13\pi}{24}.

[Argand diagram sketch]

  • Draw and label the Real (Re) and Imaginary (Im) axes.
  • Plot the four points representing the roots.
  • Each point is at a distance of r=214r = 2^{\frac{1}{4}} from the origin.
  • The points are equally spaced by π2\frac{\pi}{2} radians (90∘90^\circ), with the first root positioned slightly below the positive Real axis (at an angle of −π24-\frac{\pi}{24}).

2. Deduce the geometrical shape the roots lie on and state its cartesian equation.

Since all four roots have a constant modulus of ∣z∣=214|z| = 2^{\frac{1}{4}}, they are equidistant from the origin. Thus, the geometrical shape they lie on is a circle centered at the origin (0,0)(0,0).

The radius of the circle is r=214r = 2^{\frac{1}{4}}.
The Cartesian equation of a circle centered at the origin is x2+y2=r2x^2 + y^2 = r^2.
x2+y2=(214)2x^2 + y^2 = (2^{\frac{1}{4}})^2
x2+y2=2x^2 + y^2 = \sqrt{2}

Answer

Argand Diagram: 4 points plotted in the 4 quadrants on a circle of radius 2142^{\frac{1}{4}}, spaced evenly by π2\frac{\pi}{2} radians.

Equation: x2+y2=2x^2 + y^2 = \sqrt{2}

AI analysis — may contain mistakes.

Worked this one out on paper? Snap your steps and Gigle's AI math checker will circle the step where it went wrong.

More questions on this topic

  1. 1Solve z4=−4−4√3iz^4 = -4 - 4√3\mathrm{i}, expressing your answers in the form reiθr\mathrm{e}^{\mathrm{i}\theta}, where r>0r > 0 an…
  2. 2已知复数z1=1−3iz_{1}=1-3i,z2=3+iz_{2}=3+i,则()(\quad) A. ∣z1+z2∣=6  |z_{1}+z_{2}|=6\:\: B. z‾1−z2=−2+2i{\overline{z}}_{1}-{z}_{2}=-2+2i C. z1z2=6−8iz_{1}z_{2}=6-8i …
  3. 3Solve the equation iz5=−32i z^5 = -32, giving your roots in the form reiθre^{i\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi. Ske…
  4. 4Show that for any complex number z=eiθz = e^{i\theta}, zn+1zn=2cos⁡nθ.z^{n} + \dfrac{1}{z^{n}} = 2\cos n\theta . By taking n=1n = 1, sh…
  5. 5已知复数z=x+yi(x,y∈R)z=x+yi(x,y\in R),则()(\quad) A. z2⩾0z^{2}\geqslant 0 B. zz的虚部是yiyi C. 若z=1+2iz=1+2i,则x=1x=1,y=2y=2 D. ∣z∣=x2+y2|z|=\sqrt{x^{2}+y^{2}}…