A-Level Mathematics October/November 2024 Q3: The square roots of 6-8i can be expressed in the Cartesian form x+iy, where x and y are…

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  Let the square roots of $6-8i$ be $x+iy$, where $x, y \in \mathbb{R}$.

  Then $(x+iy)^2 = 6-8i$.

  Expanding the left-hand side: $x^2 + 2ixy + (iy)^2 = 6-8i$ $x^2 - y^2 + 2ixy = 6-8i$

  Equating the real and imaginary parts: Real part: $x^2 - y^2 = 6$ ---(1) Imaginary part: $2xy = -8$ ---(2)

  From equation (2), we can express $y$ in terms of $x$: $y = \frac{-8}{2x} = -\frac{4}{x}$

  Substitute this expression for $y$ into equation (1): $x^2 - \left(-\frac{4}{x}\right)^2 = 6$ $x^2 - \frac{16}{x^2} = 6$

  To form a quartic equation, multiply the entire equation by $x^2$: $x^4 - 16 = 6x^2$ $x^4 - 6x^2 - 16 = 0$

  This is a quadratic equation in terms of $x^2$. Let $u = x^2$: $u^2 - 6u - 16 = 0$

  Factorising the quadratic: $(u-8)(u+2) = 0$ So, $u=8$ or $u=-2$.

  This means $x^2=8$ or $x^2=-2$. Since $x$ is a real number, $x^2$ must be non-negative. Therefore, we discard $x^2=-2$.

  Using $x^2=8$: $x = \pm\sqrt{8} = \pm 2\sqrt{2}$

  Now we find the corresponding values of $y$ using $y = -\frac{4}{x}$: If $x = 2\sqrt{2}$, then $y = -\frac{4}{2\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$. If $x = -2\sqrt{2}$, then $y = -\frac{4}{-2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

  The two square roots are $2\sqrt{2} - i\sqrt{2}$ and $-2\sqrt{2} + i\sqrt{2}$.

  These can be written as $\pm(2\sqrt{2} - i\sqrt{2})$.

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  How the marks are awarded

  • M1 — The first step of setting $(x+iy)^2 = 6-8i$, expanding it, and attempting to equate the real and imaginary parts.
  • A1 — Correctly obtaining the two simultaneous equations: $x^2 - y^2 = 6$ and $2xy = -8$.
  • DM1 — Using the two equations to eliminate one variable. In the model answer, this is achieved by making $y$ the subject of the second equation ($y = -4/x$) and substituting it into the first.
  • A1 — Correctly deriving the quartic equation $x^4 - 6x^2 - 16 = 0$ after substitution and simplification.
  • A1 — Solving the quartic for real values of $x$, finding the corresponding values of $y$, and stating the two final square roots correctly in exact Cartesian form, $\pm(2\sqrt{2} - i\sqrt{2})$.

  Common mistakes

  • A sign error when equating imaginary parts, writing $2xy = 8$ instead of $2xy = -8$, which leads to an incorrect final answer.
  • An algebraic error when substituting, for example squaring $(-4/x)$ to get $-16/x^2$ instead of $+16/x^2$, resulting in an incorrect quartic equation.
  • Incorrectly pairing the values of x and y. The condition $2xy = -8$ means x and y must have opposite signs, which is a crucial check.
  • Forgetting that a square root has two solutions, and only providing one root, e.g., just $2\sqrt{2} - i\sqrt{2}$.

  Examiner tip: Master the technique of equating real and imaginary parts to convert a single complex equation into two real simultaneous equations, which can then be solved algebraically.

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