What Is The Square Root Of A Complex Number? (2 Methods To Solve)


We know how to find the square root of a real number, but what about the square root of a complex number or an imaginary number? As it turns out, we can do it, but there is a little more to it.

So, what is the square root of a complex number?  The square root of a complex number Z is a complex number S that satisfies Z = S2. Note that -S (the negative of S) is also a square root of Z. We can use polar form to find the square root of a complex number. For an imaginary number bi, the square roots are √(b/2) + i√(b/2) and -√(b/2) – i√(b/2).

Of course, we can also use algebra to solve for the square root of a complex number. However, the formulas can get a little tricky.

In this article, we’ll talk about how to find the square root of a complex number and the square root of an imaginary number, using a formula and polar form.

Let’s get going.

What Is The Square Root Of A Complex Number?

The square root of a complex number Z is another complex number S that satisfies the equation

  • Z = S2

In other words, S is the square root of Z.

We are already familiar with the square roots of real numbers:

  • If R > 0, then the square root of R is √R.
  • If R = 0, then the square root of R is 0.
  • If R < 0, then the square root of R is i√R.

Remember that a real number is really a complex number a + bi where b = 0.

The trouble in finding square roots emerges when we try to take the square root of a complex number whose imaginary part is nonzero (that is, a number a + bi where b is not zero).

complex number
A complex number a + bi has a real part “a” (graphed on the horizontal Re axis) and an imaginary part “bi” (graphed on the vertical Im axis).

Let’s start with how to find the square root of i, the imaginary unit.

What Is The Square Root Of i?

The square root of i is the complex number √(1/2) + i√(1/2).  There is a second square root of I, which is the negative of this first root: -√(1/2) – i√(1/2).

Here is one way to find the square root of i with algebra.

The square root of i is a complex number, so we’ll call it a + bi.  By the definition of a square root:

  • (a + bi)2 = i  [definition of the square root of i]
  • (a + bi)(a + bi) = i
  • a2 + abi + abi + b2i2 = i  [used FOIL to expand the product of two binomials]
  • a2 + 2abi + b2i2 = i  [combine like terms]
  • a2 + 2abi – b2 = i  [used the definition of i as the square root of -1, or i2 = -1]
  • (a2 – b2) + 2abi = 0 + 1i  [separate real and imaginary parts on each side]

Now we will write two equations: one for the real parts on each side of the equation, and one for the imaginary parts on each side of the equation:

  • a2 – b2 = 0  [set the real parts equal]
  •   2abi = 1i  [set the imaginary parts equal]

The first equation (real parts) has a difference of squares on the left side, which factors easily:

  • (a2 – b2) = 0
  • (a + b)(a – b) = 0  [factor the left side as a difference of squares]

This leaves us with a + b = 0 or a – b = 0.  Thus, either a = -b or a = b.

We will try both of these cases in the second equation:

Case 1: a = -b

  • 2abi = 1i
  • 2ab = 1
  • 2(-b)b = 1
  • -2b2 = 1
  • b2 = -1/2
  • b = +i√(1/2) or – i√(1/2)

This does not work, since b must be a real number.  Now trying the second case:

Case 2: a = b

  • 2abi = 1i
  • 2ab = 1
  • 2(b)b = 1
  • 2b2 = 1
  • b2 = 1/2
  • b = +√(1/2) or -√(1/2)

Since a = b, we have two possible solutions for the square root of i:

  • √(1/2) + i√(1/2)
  • -√(1/2) – i√(1/2)

So the imaginary unit, i, has two distinct square roots.  Note that any complex number (except zero) has two distinct square roots, which are negatives of each other.

Now, we can use the square root of i to find the square root of any imaginary number.

What Is The Square Root Of An Imaginary Number?

The square root of an imaginary number bi is the complex number √(b/2) + i√(b/2).  There is a second square root of I, which is the negative of this first root: -√(b/2) – i√(b/2).

To verify this, we can simply square this complex number by using FOIL, combining like terms, and simplifying by using i2 = -1:

  • (√(b/2) + i√(b/2))2
  • =(√(b/2) + i√(b/2))( √(b/2) + i√(b/2))
  • =b/2 + ib/2 + ib/2 + i2b/2  [use FOIL on the product of binomials]
  • =b/2 + 2bi/2 + i2b/2  [combine like terms]
  • =b/2 + 2bi/2 – b/2  [used the definition of i as the square root of -1, or i2 = -1]
  • =2bi/2  [combine like terms]
  • =bi  [cancel 2]

When we square the negative root, -√(b/2) – i√(b/2), we will get the same result, since the negatives will becomes positive when squared.

So, the two square roots of the imaginary number bi are:

  • √(b/2) + i√(b/2)
  • -√(b/2) – i√(b/2)

How To Find The Square Root Of A Complex Number (2 Methods)

What if we have a complex number a + bi, where both a and b are not zero?  In some cases, finding the square root is a little more difficult when using algebra – in that case, we might need to use a different method.

Method 1: Find The Square Root Of A Complex Number Using Algebra (FOIL & Solve A System Of Equations)

To find the square root of a complex number c + di, we can use a similar method to the one we used earlier to find the square root of i.

The square root of a + bi is a complex number, so we’ll call it c + di.  By the definition of a square root:

  • (c + di)2 = a + bi  [definition of the square root of i]
  • (c + di)(c + di) = a + bi
  • c2 + cdi + cdi + d2i2 = a + bi  [used FOIL to expand the product of two binomials]
  • c2 + 2cdi + b2i2 = a + bi  [combine like terms]
  • c2 + 2cdi – b2 = a + bi  [used the definition of i as the square root of -1, or i2 = -1]
  • (c2 – d2) + 2cdi = a + bi  [separate real and imaginary parts on each side]

Now we will write two equations: one for the real parts on each side of the equation, and one for the imaginary parts on each side of the equation:

  • c2 – d2 = a  [set the real parts equal]
  • 2cdi = bi  [set the imaginary parts equal]

Solving the second equation for c, we get c = b/2d.  Using this in the first equation, we get:

  • c2 – d2 = a 
  • (b/2d)2 – d2 = a  [c = b/2d]
  • b2/4d2 – d2 = a
  • b2 – 4d4 = 4ad2
  • 4d4 + 4ad2 – b2 = 0  [put all terms on one side]
  • 4x2 + 4ax – b2 = 0  [make the change of variables x = d2]

Now we need to solve this quadratic equation in x, with coefficients A = 4, B = 4a, and C = -b2.  Using the quadratic formula gives us:

x = -1/2(a + √(a2 + b2)) and x = -1/2(a – √(a2 + b2))

Since x = d2, we would need to take the square root of each of these expressions to find d.

Since d is a real number, we need to take the square root of the positive number, which is x = -1/2(a – √(a2 + b2)).

So, d = √(-1/2(a – √(a2 + b2))) and d = -√(-1/2(a – √(a2 + b2))).

We can also find c by using the equation c = b/2d.

Example: Find The Square Root of 4 + 3i.

In this case, we have the complex number 4 + 3i with a = 4 and b = 3.

Then:

  • d = √(-1/2(a – √(a2 + b2)))
  • d = √(-1/2(4 – √(42 + 32)))
  • d = √(-1/2(4 – √(16 + 9)))
  • d = √(-1/2(4 – √(25)))
  • d = √(-1/2(-1))
  • d = √(1/2)

We can also find c with:

  • c = b/2d
  • c = 3 / 2(√(1/2))
  • c = 3/√2

So, the square root of 4 + 3i is c + di = 3/√2 + i√(1/2).

We can verify this by squaring 3/√2 + i√(1/2) to get 3 + 4i.

The negative of this complex number, -3/√2 – i√(1/2), is also a square root of 3 + 4i.

Method 2: Find The Square Root Of A Complex Number Using Polar Form (Modulus & Argument)

To find the square root of a complex number, it is much easier to convert to polar form first and then take the square root.

complex number with angles
The polar form of a complex number helps us to find square roots more easily. The complex number ReiA (in polar form) is equivalent to R(cos(A) + isin(A)), or Rcos(A) + iRsin(A).

To convert a complex number a + bi to polar form, we need to calculate both the modulus and the argument.

The modulus R (a real number) is given by the equation

  • R = √(a2 + b2)

The argument A (an angle) is given by the equation

  • A = tan-1(b/a)

The polar form of the complex number a + bi is

  • ReiA

where R is the modulus √(a2 + b2) and A is the argument tan-1(b/a).  In other words,

  • a + bi = ReiA

Note: remember that a and A are not necessarily the same number!

Once we have polar form, we can find the square root.  We use half of A as our new angle and take the square root of the modulus R as our new modulus.

So, the square root of a + bi = ReiA is:

  • (√R)*eiA/2

The negative of this number, -(√R)*eiA/2, is also a square root of a + bi.

Let’s look at an example of how this works.

Example: Find The Square Root of 4 + 3i Using Polar Form

For the complex number 4 + 3i, we have a = 4 and b = 3.

The modulus R is:

  • R = √(a2 + b2)
  • R = √(42 + 32)
  • R = √(16 + 9)
  • R = √(25)
  • R = 5

The argument A is:

  • A = tan-1(b/a)
  • A = tan-1(3/4)
  • A ~0.6435 radians

Then the polar form of the complex number 4 + 3i is:

  • ReiA
  • =5e0.6435i

To take the square root, we take the square root of the modulus and half of the argument to get:

  • (√5)* e0.32175i
  • 2.2121e0.1024i

Converting this back to a complex number, we get approximately 2.1213 + 0.7071i, which corresponds to our solution from earlier: 3/√2 + i√(1/2).

Conclusion

Now you know how to find the square root of an imaginary number and the square root of a complex number.  You also know how to convert from a complex number to polar form.

You can learn more about square roots in my article here.

You can also learn all about what square roots are used for in my article here.

You can learn how to do square roots by hand in my article here.

I hope you found this article helpful.  If so, please share it with someone who can use the information.

Don’t forget to subscribe to my YouTube channel & get updates on new math videos!

~Jonathon

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