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What Is The Square Root Formula Of A Complex Number?

Published Aug 29, 2025 9 min read
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For a complex number z=a+biz equals a plus b i 𝑧=π‘Ž+𝑏𝑖 , the square root formula is:a+bi=Β±(a2+b2+a2+iβ‹…sgn(b)a2+b2βˆ’a2)the square root of a plus b i end-root equals plus or minus open paren the square root of the fraction with numerator the square root of a squared plus b squared end-root plus a and denominator 2 end-fraction end-root plus i center dot sgn open paren b close paren the square root of the fraction with numerator the square root of a squared plus b squared end-root minus a and denominator 2 end-fraction end-root close paren

π‘Ž+π‘π‘–βˆš=Β±βŽβŽœβŽœβŽ›π‘Ž2+𝑏2√+π‘Ž2+𝑖⋅sgn(𝑏)π‘Ž2+𝑏2βˆšβˆ’π‘Ž2⎠⎟⎟⎞

.

This formula provides two square roots, which are negations of each other. The sgn(b)sgn open paren b close paren

sgn(𝑏)

term is the sign function, which ensures the correct sign for the imaginary part of the root:

  • If b>0b is greater than 0

    𝑏>0

    , then sgn(b)=1sgn open paren b close paren equals 1

    sgn(𝑏)=1

    , and the real and imaginary parts of the root have the same sign.

  • If b<0b is less than 0

    𝑏<0

    , then sgn(b)=-1sgn open paren b close paren equals negative 1

    sgn(𝑏)=βˆ’1

    , and the real and imaginary parts have opposite signs.

Derivation of the formula

To understand where this formula comes from, we derive it using basic algebraic principles. Let's assume the square root of a+bia plus b i

π‘Ž+𝑏𝑖

is another complex number x+yix plus y i

π‘₯+𝑦𝑖

.a+bi=x+yithe square root of a plus b i end-root equals x plus y i

π‘Ž+π‘π‘–βˆš=π‘₯+𝑦𝑖

  1. **Square both sides:**a+bi=(x+yi)2a plus b i equals open paren x plus y i close paren squared

    π‘Ž+𝑏𝑖=(π‘₯+𝑦𝑖)2

    a+bi=(x2βˆ’y2)+2xyia plus b i equals open paren x squared minus y squared close paren plus 2 x y i

    π‘Ž+𝑏𝑖=(π‘₯2βˆ’π‘¦2)+2π‘₯𝑦𝑖

  2. **Equate the real and imaginary parts:**By comparing the real and imaginary parts of the equation, we get a system of two equations with two variables, xx

    π‘₯

    and yy

    𝑦

    :

    • a=x2βˆ’y2a equals x squared minus y squared

      π‘Ž=π‘₯2βˆ’π‘¦2

      (Equation 1)

    • b=2xyb equals 2 x y

      𝑏=2π‘₯𝑦

      (Equation 2)

  3. **Use the modulus to find another relationship:**We can also find the modulus (magnitude) of the square root. The modulus of z=a+biz equals a plus b i

    𝑧=π‘Ž+𝑏𝑖

    is |z|=a2+b2the absolute value of z end-absolute-value equals the square root of a squared plus b squared end-root

    |𝑧|=π‘Ž2+𝑏2√

    . The modulus of the square root is |z|the square root of the absolute value of z end-absolute-value end-root

    |𝑧|√

    .|x+yi|2=|a+bi|the absolute value of x plus y i end-absolute-value squared equals the absolute value of a plus b i end-absolute-value

    |π‘₯+𝑦𝑖|2=|π‘Ž+𝑏𝑖|

    x2+y2=a2+b2x squared plus y squared equals the square root of a squared plus b squared end-root

    π‘₯2+𝑦2=π‘Ž2+𝑏2√

    (Equation 3)

  4. Solve the system of equations:

    • Add Equation 1 and Equation 3:(x2βˆ’y2)+(x2+y2)=a+a2+b2open paren x squared minus y squared close paren plus open paren x squared plus y squared close paren equals a plus the square root of a squared plus b squared end-root

      (π‘₯2βˆ’π‘¦2)+(π‘₯2+𝑦2)=π‘Ž+π‘Ž2+𝑏2√

      2x2=a+a2+b22 x squared equals a plus the square root of a squared plus b squared end-root

      2π‘₯2=π‘Ž+π‘Ž2+𝑏2√

      x2=a2+b2+a2x squared equals the fraction with numerator the square root of a squared plus b squared end-root plus a and denominator 2 end-fraction

      π‘₯2=π‘Ž2+𝑏2√+π‘Ž2

      x=Β±a2+b2+a2x equals plus or minus the square root of the fraction with numerator the square root of a squared plus b squared end-root plus a and denominator 2 end-fraction end-root

      π‘₯=Β±π‘Ž2+𝑏2√+π‘Ž2

    • Subtract Equation 1 from Equation 3:(x2+y2)βˆ’(x2βˆ’y2)=a2+b2βˆ’aopen paren x squared plus y squared close paren minus open paren x squared minus y squared close paren equals the square root of a squared plus b squared end-root minus a

      (π‘₯2+𝑦2)βˆ’(π‘₯2βˆ’π‘¦2)=π‘Ž2+𝑏2βˆšβˆ’π‘Ž

      2y2=a2+b2βˆ’a2 y squared equals the square root of a squared plus b squared end-root minus a

      2𝑦2=π‘Ž2+𝑏2βˆšβˆ’π‘Ž

      y2=a2+b2βˆ’a2y squared equals the fraction with numerator the square root of a squared plus b squared end-root minus a and denominator 2 end-fraction

      𝑦2=π‘Ž2+𝑏2βˆšβˆ’π‘Ž2

      y=Β±a2+b2βˆ’a2y equals plus or minus the square root of the fraction with numerator the square root of a squared plus b squared end-root minus a and denominator 2 end-fraction end-root

      𝑦=Β±π‘Ž2+𝑏2βˆšβˆ’π‘Ž2

  5. **Determine the signs of xx

    π‘₯

    and yy

    𝑦

    :**The original equation b=2xyb equals 2 x y

    𝑏=2π‘₯𝑦

    dictates the signs of xx

    π‘₯

    and yy

    𝑦

    .

    • If bb

      𝑏

      is positive, xx

      π‘₯

      and yy

      𝑦

      must have the same sign (both positive or both negative).

    • If bb

      𝑏

      is negative, xx

      π‘₯

      and yy

      𝑦

      must have opposite signs.The sgn(b)sgn open paren b close paren

      sgn(𝑏)

      term handles this automatically by being positive when b>0b is greater than 0

      𝑏>0

      and negative when b<0b is less than 0

      𝑏<0

      .

  6. **Combine the results:**Combining the expressions for xx

    π‘₯

    and yy

    𝑦

    along with the sign rule gives the final formula for the two square roots:a+bi=Β±(a2+b2+a2+iβ‹…sgn(b)a2+b2βˆ’a2)the square root of a plus b i end-root equals plus or minus open paren the square root of the fraction with numerator the square root of a squared plus b squared end-root plus a and denominator 2 end-fraction end-root plus i center dot sgn open paren b close paren the square root of the fraction with numerator the square root of a squared plus b squared end-root minus a and denominator 2 end-fraction end-root close paren

    π‘Ž+π‘π‘–βˆš=Β±βŽβŽœβŽœβŽ›π‘Ž2+𝑏2√+π‘Ž2+𝑖⋅sgn(𝑏)π‘Ž2+𝑏2βˆšβˆ’π‘Ž2⎠⎟⎟⎞

An alternative method: The polar form

Finding the square root of a complex number can be much simpler using the polar form, which represents a complex number by its modulus (distance from origin) and argument (angle).

  1. **Convert to polar form:**A complex number z=a+biz equals a plus b i

    𝑧=π‘Ž+𝑏𝑖

    can be written as z=r(cosΞΈ+isinΞΈ)z equals r open paren cosine theta plus i sine theta close paren

    𝑧=π‘Ÿ(cosπœƒ+𝑖sinπœƒ)

    , where:

    • r=a2+b2r equals the square root of a squared plus b squared end-root

      π‘Ÿ=π‘Ž2+𝑏2√

      is the modulus.

    • ΞΈ=arctan(b/a)theta equals arc tangent open paren b / a close paren

      πœƒ=arctan(𝑏/π‘Ž)

      is the argument (adjusted for the correct quadrant).

  2. **Apply De Moivre's Theorem for roots:**For a complex number z=r(cosΞΈ+isinΞΈ)z equals r open paren cosine theta plus i sine theta close paren

    𝑧=π‘Ÿ(cosπœƒ+𝑖sinπœƒ)

    , its square roots are given by:z=z1/2=r1/2(cos(ΞΈ+2kΟ€2)+isin(ΞΈ+2kΟ€2))the square root of z end-root equals z raised to the 1 / 2 power equals r raised to the 1 / 2 power open paren cosine open paren the fraction with numerator theta plus 2 k pi and denominator 2 end-fraction close paren plus i sine open paren the fraction with numerator theta plus 2 k pi and denominator 2 end-fraction close paren close paren

    π‘§βˆš=𝑧1/2=π‘Ÿ1/2cosπœƒ+2π‘˜πœ‹2+𝑖sinπœƒ+2π‘˜πœ‹2

    for k=0,1k equals 0 comma 1

    π‘˜=0,1

    .

  3. Find the two roots:

    • For k=0k equals 0

      π‘˜=0

      , the first root is z1=r(cos(ΞΈ2)+isin(ΞΈ2))z sub 1 equals the square root of r end-root open paren cosine open paren the fraction with numerator theta and denominator 2 end-fraction close paren plus i sine open paren the fraction with numerator theta and denominator 2 end-fraction close paren close paren

      𝑧1=π‘Ÿβˆšcosπœƒ2+𝑖sinπœƒ2

      .

    • For k=1k equals 1

      π‘˜=1

      , the second root is z2=r(cos(ΞΈ+2Ο€2)+isin(ΞΈ+2Ο€2))=r(cos(ΞΈ2+Ο€)+isin(ΞΈ2+Ο€))z sub 2 equals the square root of r end-root open paren cosine open paren the fraction with numerator theta plus 2 pi and denominator 2 end-fraction close paren plus i sine open paren the fraction with numerator theta plus 2 pi and denominator 2 end-fraction close paren close paren equals the square root of r end-root open paren cosine open paren the fraction with numerator theta and denominator 2 end-fraction plus pi close paren plus i sine open paren the fraction with numerator theta and denominator 2 end-fraction plus pi close paren close paren

      𝑧2=π‘Ÿβˆšcosπœƒ+2πœ‹2+𝑖sinπœƒ+2πœ‹2=π‘Ÿβˆšcosπœƒ2+πœ‹+𝑖sinπœƒ2+πœ‹

      .

    Since cos(Ο•+Ο€)=βˆ’cos(Ο•)cosine open paren phi plus pi close paren equals negative cosine open paren phi close paren

    cos(πœ™+πœ‹)=βˆ’cos(πœ™)

    and sin(Ο•+Ο€)=βˆ’sin(Ο•)sine open paren phi plus pi close paren equals negative sine open paren phi close paren

    sin(πœ™+πœ‹)=βˆ’sin(πœ™)

    , the second root is simply the negative of the first root: z2=βˆ’z1z sub 2 equals negative z sub 1

    𝑧2=βˆ’π‘§1

    .

Example: Find the square root of 3+4i3 plus 4 i

3+4𝑖

We can solve this using both methods.

Method 1: Using the Cartesian formula

  • Given a=3a equals 3

    π‘Ž=3

    and b=4b equals 4

    𝑏=4

    .

  • Calculate the modulus: a2+b2=32+42=9+16=25=5the square root of a squared plus b squared end-root equals the square root of 3 squared plus 4 squared end-root equals the square root of 9 plus 16 end-root equals the square root of 25 end-root equals 5

    π‘Ž2+𝑏2√=32+42√=9+16√=25√=5

    .

  • Since b>0b is greater than 0

    𝑏>0

    , we use the positive sign for both the real and imaginary parts of the formula.

  • **Real part:**a2+b2+a2=5+32=82=4=2the square root of the fraction with numerator the square root of a squared plus b squared end-root plus a and denominator 2 end-fraction end-root equals the square root of the fraction with numerator 5 plus 3 and denominator 2 end-fraction end-root equals the square root of eight-halves end-root equals the square root of 4 end-root equals 2

    π‘Ž2+𝑏2√+π‘Ž2=5+32=82=4√=2

    .

  • **Imaginary part:**a2+b2βˆ’a2=5βˆ’32=22=1=1the square root of the fraction with numerator the square root of a squared plus b squared end-root minus a and denominator 2 end-fraction end-root equals the square root of the fraction with numerator 5 minus 3 and denominator 2 end-fraction end-root equals the square root of two-halves end-root equals the square root of 1 end-root equals 1

    π‘Ž2+𝑏2βˆšβˆ’π‘Ž2=5βˆ’32=22=1√=1

    .

  • The two square roots are Β±(2+i)plus or minus open paren 2 plus i close paren

    Β±(2+𝑖)

    .

Method 2: Using the Polar form

  • **Step 1: Convert to polar form.**r=32+42=5r equals the square root of 3 squared plus 4 squared end-root equals 5

    π‘Ÿ=32+42√=5

    .ΞΈ=arctan(4/3)β‰ˆ53.13∘theta equals arc tangent open paren 4 / 3 close paren is approximately equal to 53.13 raised to the composed with power

    πœƒ=arctan(4/3)β‰ˆ53.13∘

    .So 3+4i=5(cos(53.13∘)+isin(53.13∘))3 plus 4 i equals 5 open paren cosine open paren 53.13 raised to the composed with power close paren plus i sine open paren 53.13 raised to the composed with power close paren close paren

    3+4𝑖=5(cos(53.13∘)+𝑖sin(53.13∘))

    .

  • **Step 2: Apply De Moivre's Theorem.**3+4i=5(cos(53.13∘+360∘k2)+isin(53.13∘+360∘k2))the square root of 3 plus 4 i end-root equals the square root of 5 end-root open paren cosine open paren the fraction with numerator 53.13 raised to the composed with power plus 360 raised to the composed with power k and denominator 2 end-fraction close paren plus i sine open paren the fraction with numerator 53.13 raised to the composed with power plus 360 raised to the composed with power k and denominator 2 end-fraction close paren close paren

    3+4π‘–βˆš=5√cos53.13∘+360βˆ˜π‘˜2+𝑖sin53.13∘+360βˆ˜π‘˜2

    .

  • Step 3: Find the roots for k=0 and k=1.

    • For k=0k equals 0

      π‘˜=0

      : 5(cos(26.565∘)+isin(26.565∘))β‰ˆ5(0.894+0.447i)β‰ˆ2(0.894)+2(0.447)iβ‰ˆ1.788+0.894ithe square root of 5 end-root open paren cosine open paren 26.565 raised to the composed with power close paren plus i sine open paren 26.565 raised to the composed with power close paren close paren is approximately equal to the square root of 5 end-root open paren 0.894 plus 0.447 i close paren is approximately equal to 2 open paren 0.894 close paren plus 2 open paren 0.447 close paren i is approximately equal to 1.788 plus 0.894 i

      5√(cos(26.565∘)+𝑖sin(26.565∘))β‰ˆ5√(0.894+0.447𝑖)β‰ˆ2(0.894)+2(0.447)π‘–β‰ˆ1.788+0.894𝑖

      .

    • This seems different, but that's because 2+i2 plus i

      2+𝑖

      is the simplified form.

    • Check: (2+i)2=4+4i+i2=4+4iβˆ’1=3+4iopen paren 2 plus i close paren squared equals 4 plus 4 i plus i squared equals 4 plus 4 i minus 1 equals 3 plus 4 i

      (2+𝑖)2=4+4𝑖+𝑖2=4+4π‘–βˆ’1=3+4𝑖

      . Our answer is correct.

    The small discrepancy comes from rounding the angle. The Cartesian formula is often more precise for direct calculation.

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