MYTUTOR SUBJECT ANSWERS

496 views

How do you find the square roots of a complex number?

Every complex number has complex square roots. However since we don't know how to deal with expressions such as √i we need to follow a specific method to find the square roots of a complex number.

Let's consider the complex number 21-20i.

We know that all square roots of this number will satisfy the equation 21-20i=x2 by definition of a square root.

We also know that x can be expressed as a+bi (where a and b are real) since the square roots of a complex number are always complex.

So 21-20i=(a+bi)2.

The natural step to take here is the mulitply out the term on the right-hand side.

This gives 21-20i=a2+(2ab)i+(b2)i2.

As i2=-1 by definition of i, this equation can be rearranged to give 21-20i=(a2-b2)+(2ab)i.

Now both sides of the equation are in the same form.

Let's compare coeffiecients to obtain two equations in a and b.

First, let's compare the real parts of the equation.

We have a2-b2=21 (call this equation 1).

Next, let's compare the imaginary parts of the equation (the coefficients of i).

We have 2ab=-20 (call this equation 2).

We now have two equations in two unknowns. We can solve these simultaneous equations for a and b.

Firstly, we can make b the subject of equation 2 by dividing both sides by 2a.

We have b=-10/a.

Now substitute this expression for b into equation 1.

We have a2-(-10/a)2=21.

Some simplification and factorisation of this equation gives us (a2+4)(a2-25)=0, a quadratic in disguise.

So either a2=-4 or a2=25.

We have assumed a to be real so a2=-4 has no solutions of interest to us.

This means our solutions are a=5 and a=-5.

Substitute each a value into our earlier expression for b.

This means that when a=5, b=-2 and when a=-5, b=2.

So, putting a and b back into the context of the question, we have two solutions: 5-2i and -5+2i.

Jacob G. A Level Maths tutor, A Level Further Mathematics  tutor, GCS...

1 year ago

Answered by Jacob, an A Level Further Mathematics tutor with MyTutor

Still stuck? Get one-to-one help from a personally interviewed subject specialist

68 SUBJECT SPECIALISTS

£20 /hr

Ben H.

Degree: Mathematics (Masters) - Bristol University

Subjects offered: Further Mathematics , Maths

Further Mathematics
Maths

“Who am I? I am currently studying for an undergraudate and masters degree in Mathematics at the University of Bristol. Maths has always been my main interest and I have always enjoyed the challenge of teaching myself new content, for ...”

MyTutor guarantee

£20 /hr

Ben J.

Degree: Mathematics (Masters) - Durham University

Subjects offered: Further Mathematics , Maths

Further Mathematics
Maths

“Friendly tutor with perfect academic record happy to help you enjoy Maths and get the grades you need! :)”

£20 /hr

Maisie J.

Degree: Mathematics (G100) (Bachelors) - Warwick University

Subjects offered: Further Mathematics , Maths

Further Mathematics
Maths

“Hi! My name is Maisie and I'm studying Mathematics at Warwick University. My aim is to share my passion of maths with my students and to boost my students' self-confidence and self-esteem.  I am friendly, patient and I will adapt the ...”

MyTutor guarantee

About the author

Jacob G.

Currently unavailable: for regular students

Degree: Mathematics (Bachelors) - Warwick University

Subjects offered: Further Mathematics , Maths+ 2 more

Further Mathematics
Maths
History
English

“About Me: Hi, I'm Jacob and I'm a second-year student of Mathematics at the University of Warwick. My main subject isMaths and I averaged 98.5% over my Maths and Further Maths A-level exams. I also maintain an interest inEnglish and Hi...”

MyTutor guarantee

You may also like...

Posts by Jacob

How do you 'rationalise the denominator'?

How do you find the square roots of a complex number?

Other A Level Further Mathematics questions

How do you calculate the derivative of cos inverse x?

What is sin(x)/x for x =0?

The roots of the equation z^3 + 2z^2 +3z - 4 = 0, are a, b and c . Show that a^2 + b^2 +c^2 = -2

Prove by induction that 11^n - 6 is divisible by 5 for all positive integer n.

View A Level Further Mathematics tutors

Cookies:

We use cookies to improve our service. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok