- From: Lee Armstrong
- Date: 3 March 1999
*Subject: Solve this quadratic with complex coefficients*
Can you show me how to solve the equation None of us could do it in class - even the teacher got stuck! (I am doing Further Maths A-level) |

*Preliminary comment: This topic of complex numbers occurs in the "further" pure maths
syllabus at A-level. Complex numbers are also dealt with in some higher engineering courses.
Some textbooks use i for the imaginary value sqrt(-1), some use j.
In this question, j^{2} = -1*

Let's look closely at your equation.

zThis is basically a quadratic, in the usual form^{2}+ (-5 + j)z + 6 - 2j = 0

azwhere a,b,c are coefficients. The only special thing here is that they are^{2}+ bz + c = 0

a = 1Substituting these values into the well-known equation for the roots of a quadratic, we get

b = -5 + j

c = 6 - 2j

At this stage we pause - what is the square root of -2j? Finding the square root of a complex number is fairly tricky (but it is an important technique in Further Pure Maths A-level papers). We will explain at the end of this page how to do it.

For now, just accept that:

**sqrt(-2j) = 1 - j OR -1 + j**

Using this "fact", we proceed to the solution:

How do we check these solutions are correct? The long way is to substitute them for z in the
original equation, and verify that the left hand side does simplify to 0. A neater way is to
recall the standard result for roots of quadratics:

Sum of roots = -b/aI suggest you carry out each of these checks for practice.

Product of roots = c/a

EXAMPLE

Find the square root of 7 + 24j

METHOD

We assume that the answer will be another complex number. Call it **u + vj**, where u and v
are **real** numbers. Thus:

u + vj = sqrt(7 + 24j)Comparing real and imaginary parts gives us two simultaneous equations in u and v:

(u + vj)^{2}= 7 + 24j

u^{2}+ 2uvj + v^{2}j^{2}= 7 + 24j

u^{2}+ 2uvj - v^{2}= 7 + 24j

uFrom (2) we have^{2}- v^{2}= 7 ....(1)

2uv = 24 ....(2)

v = 12/uSubstituting for u in (1) gives:

uThe second option is not possible, because u must be real (see statement at the beginning).^{2}- 144/u^{2}= 7

u^{4}- 144 = 7u^{2}

u^{4}- 7u^{2}- 144 = 0

(u^{2}- 16)(u^{2}+ 9) = 0

u^{2}= 16 OR u^{2}= -9

uSubstituting these values for u back in (2) gives^{2}= 16

u = 4 OR -4

when u = 4 , v = 3Therefore the square roots of 7 + 24j are

when u = -4 , v = -3

NOTE: Just as with real numbers, there are two square roots of a complex number, of opposite signs.4 + 3j OR -4 - 3j

EXERCISE:

Use the method above to show that sqrt(-2j) = 1 - j OR -1 + j

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