Square roots followed by a quadratic shift

(i) The roots of w2=4iw^2=4i are w1,w2w_1,w_2. Find w1,w2w_1,w_2 in Cartesian form x+iyx+iy, showing your working. marks

(ii) Hence, or otherwise, find in exact Cartesian form the roots v1,v2v_1,v_2 of v2−10v+(25−i)=0v^2-10v+(25-i)=0. marks

[2024/RVHS/Prelim/I/5b]

Total marks: 6

Solution

(i)

Square root of a complex number

Find the roots w1,w2w_1,w_2 of w2=4iw^2=4i in Cartesian form x+iyx+iy, showing your working.

Explore this idea

Hint

  1. Write w=x+iyw=x+iy and expand w2w^2.

  2. Equate real and imaginary parts and solve.

Answer · (i)

w=±2(1+i)w=\pm\sqrt2(1+i).

Explore the idea

Squaring doubles the angle

Turn uu around the circle of radius 22. Its square u2u^2 turns twice as fast and has modulus 44. Stop where u2u^2 hits 4i4i.

Argument of u20°
Try an example
u
1.879 + 0.684i
u²
3.064 + 2.571i
distance from u² to 4i
3.38
  • Moves as the slider changes; a faint path shows where it goes
Key idea

Every non-zero complex number has two square roots at opposite points of a circle: half the angle, square root of the modulus.

Ideas to try
  1. Go to 45∘45^\circ: u=2(1+i)≈1.41+1.41iu=\sqrt2(1+i)\approx1.41+1.41i and u2=4iu^2=4i.
  2. Go to 225∘225^\circ for the other root. Its angle is exactly 180∘180^\circ further, because doubling it adds a full turn.
(ii)

Reduce to part (i) by completing the square

Hence find the roots v1,v2v_1,v_2 of v2−10v+(25−i)=0v^2-10v+(25-i)=0 in exact Cartesian form.

Explore this idea

Hint

  1. Recognise v2−10v+25=(v−5)2v^2-10v+25=(v-5)^2.

  2. Set w=2(v−5)w=2(v-5) or use the square roots of ii directly.

Answer · (ii)

v=5±(1+i)/2v=5\pm(1+i)/\sqrt2.

Explore the idea

The roots sit on a unit circle around 5

Drag to look around the Argand plane. Real parts run along the horizontal axis and imaginary parts along the vertical axis.

Reset the controls
Key idea

The shift by 55 moves the picture of (v−5)2=i(v-5)^2=i: the same two roots as for w2=iw^2=i, centred at 55.

Ideas to try
  1. ∣v−5∣2=∣i∣=1|v-5|^2=|i|=1, so both roots are 11 unit from 55, on the unit circle centred at 55.
  2. The roots are opposite each other, with 55 as the midpoint.
Takeaway

Recognise the square

A quadratic that completes to (v−h)2=k(v-h)^2=k has roots h±kh\pm\sqrt k. The square roots of kk come from solving w2=kw^2=k in parts.