A-Level Mathematics May/June 2025 Q5: The square roots of -1-4√5 i can be expressed in the Cartesian form x+iy, where x and y…

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  Let the square root of $-1-4\sqrt{5}i$ be $x+iy$, where $x$ and $y$ are real numbers.

  $(x+iy)^2 = -1-4\sqrt{5}i$

  Expanding the left side: $x^2 + 2xyi + (iy)^2 = -1-4\sqrt{5}i$ $x^2 - y^2 + 2xyi = -1-4\sqrt{5}i$

  Equating the real and imaginary parts: Real parts: $x^2 - y^2 = -1$ (1) Imaginary parts: $2xy = -4\sqrt{5}$ (2)

  From equation (2), we can express $y$ in terms of $x$: $y = \frac{-4\sqrt{5}}{2x} = -\frac{2\sqrt{5}}{x}$

  Substitute this expression for $y$ into equation (1): $x^2 - \left(-\frac{2\sqrt{5}}{x}\right)^2 = -1$ $x^2 - \frac{4 \times 5}{x^2} = -1$ $x^2 - \frac{20}{x^2} = -1$

  To form a quartic equation, multiply the entire equation by $x^2$: $x^4 - 20 = -x^2$ $x^4 + x^2 - 20 = 0$

  This is a quadratic equation in terms of $x^2$. Let $u = x^2$. $u^2 + u - 20 = 0$ Factorising the quadratic: $(u+5)(u-4) = 0$ So, $u = -5$ or $u = 4$.

  Substituting back $u=x^2$: $x^2 = -5$ or $x^2 = 4$. Since $x$ is a real number, $x^2$ cannot be negative. We discard $x^2 = -5$. Therefore, $x^2 = 4$, which gives $x = 2$ or $x = -2$.

  Now we find the corresponding values of $y$ using $y = -\frac{2\sqrt{5}}{x}$: Case 1: If $x=2$, then $y = -\frac{2\sqrt{5}}{2} = -\sqrt{5}$. The first root is $2 - i\sqrt{5}$.

  Case 2: If $x=-2$, then $y = -\frac{2\sqrt{5}}{-2} = \sqrt{5}$. The second root is $-2 + i\sqrt{5}$.

  The two square roots are $2 - i\sqrt{5}$ and $-2 + i\sqrt{5}$. These can be written as $\pm(2 - i\sqrt{5})$.

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  How the marks are awarded

  • M1 — Awarded for the initial step of setting $(x+iy)^2 = -1-4\sqrt{5}i$ and attempting to expand the left side to equate real and imaginary parts.
  • A1 — Awarded for correctly obtaining the two simultaneous equations by equating the real and imaginary parts: $x^2 - y^2 = -1$ and $2xy = -4\sqrt{5}$.
  • DM1 — Awarded for the dependent method of eliminating one variable. In the model answer, this is achieved by making $y$ the subject of the second equation and substituting it into the first.
  • A1 — Awarded for correctly deriving the quartic equation $x^4 + x^2 - 20 = 0$ (or the equivalent in $y$). This must be fully correct with no sign errors.
  • A1 — Awarded for correctly solving the quartic to find both exact roots, $2 - i\sqrt{5}$ and $-2 + i\sqrt{5}$, and providing no extra incorrect solutions.

  Common mistakes

  • Sign errors when equating parts, such as writing $2xy = 4\sqrt{5}$, which leads to an incorrect quartic equation and final answers.
  • Incorrectly solving the quartic equation, for example by trying to use the solution $x^2 = -5$ which is invalid as $x$ must be real, or by only finding one value for $x$ (e.g., $x=2$) and thus only one of the two square roots.
  • Arithmetic errors during substitution, such as squaring $(-2\sqrt{5})$ to get $10$ or $-20$ instead of the correct $20$.
  • Forgetting to find the second root after finding the first. The question asks for 'square roots' (plural), so both must be stated for the final mark.

  Examiner tip: When dealing with equations of complex numbers, equating the real and imaginary parts separately is a powerful strategy to create a system of solvable real equations.

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