Multiplication of Complex Numbers

Add yours
Multiplying Complex Numbers

Tip: This isn't the place to ask a question because the teacher can't reply.

Key Questions

  • First, complex numbers can come in a variety of forms!

    Ex: multiply #3i*-4i =#

    Remember, with multiplication you can rearrange the order (called the Commutative Property):

    #3*-4*i*i =-12i^2#

    ... and then always substitute -1 for #i^2#:

    #-12*-1 = 12#

    Ex: the numbers might come in a radical form:

    #sqrt(-3)*4sqrt(-12) =#

    You should always "factor" out the imaginary part from the square roots like this:

    #sqrt(-1)sqrt(3)*4*sqrt(-1)sqrt(4)sqrt(3) =#

    and simplify again:

    #=i*4*sqrt(3)*sqrt(3)*sqrt(4)#
    #=i*4*3*2 = 24i#

    Ex: what about the Distributive Property? #3i(4i - 6) =#

    #=12i^2- 18i#
    #=12(-1) - 18i#
    #= -12 - 18i#

    And last but not least, a pair of binomials in a + bi form:

    Ex: (3 - 2i)(4 + i) =

    =12 + 3i - 8i - #2i^2#
    = 12 - 2(-1) + 3i - 8i
    = 12 + 2 - 5i
    = 14 - 5i

  • Let #z_1# and #z_2# be two complex numbers.

    By rewriting in exponential form,

    #{(z_1=r_1e^{i theta_1}),(z_2=r_2 e^{i theta_2}):}#

    So,

    #z_1 cdot z_2 =r_1e^{i theta_1}cdot r_2 e^{i theta_2} =(r_1 cdot r_2)e^{i(theta_1+theta_2)}#

    Hence, the product of two complex numbers can be geometrically interpreted as the combination of the product of their absolute values (#r_1 cdot r_2#) and the sum of their angles (#theta_1+theta_2#) as shown below.

    enter image source here


    I hope that this was clear.

  • In trigonometric form, a complex number looks like this:
    #a + bi = c*cis(theta)#
    where #a#, #b# and #c# are scalars.

    Let two complex numbers:
    #-> k_(1) = c_(1)*cis(alpha)#
    #-> k_(2) = c_(2)*cis(beta)#
    #k_(1)*k_(2) = c_(1) * c_(2) * cis(alpha) * cis(beta) =#
    #= c_(1) * c_(2) * (cos(alpha) + i*sin(alpha)) * (cos(beta) + i*sin(beta))#

    This product will end up leading to the expression
    #k_(1)*k_(2) =#
    #= c_(1)*c_(2)*(cos(alpha + beta) + i*sin(alpha + beta)) =#
    #= c_(1)*c_(2)*cis(alpha+beta)#

    By analyzing the steps above, we can infer that, for having used generic terms #c_(1)#, #c_(2)#, #alpha# and #beta#, the formula of the product of two complex numbers in trigonometric form is:
    #(c_(1) * cis(alpha)) * (c_(2) * cis(beta)) = c_(1)*c_(2)*cis(alpha+beta)#

    Hope it helps.

  • To explain this, I will name two generic complex.
    #c_1 = a*cis(alpha)# and #c_2 = b*cis(beta)#

    The product between #c_1# and #c_2# is:
    #ab*cis(alpha)cis(beta) =#
    #ab*(cos(alpha)+isin(alpha)) (cos(beta)+isin(beta)) =#
    #ab*({cos(alpha)cos(beta)-sin(alpha)sin(beta)} +#
    #{i(sin(alpha)sin(beta)+cos(alpha)sin(beta)}) =#
    #ab*{cos(a+b)+isin(a+b)}#//

    Therefore, we can assume that the product of the two complex numbers #c_1# and #c_2# can be generaly given by the form above.

    Ex.:
    #(2*cis(pi)) * (3*cis(2pi)) = 6*cis(3pi) = 6*cis(pi)#

    Hope it helps.

  • This key question hasn't been answered yet. Answer question

Questions

  • Sam F. answered · 3 months ago