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Division of Complex Numbers

Key Questions

  • Answer:

    By definition if #z=a+bi# is a complex number, then his conjugate is #barz=a-bi#

    Explanation:

    A conjugate of a complex number is the other complex number with the same real part and opposite imaginary part

  • You can do it by multiplying the numerator and the denominator by the complex conjugate of the denominator.

    #1/{a+bi}=1/{a+bi}cdot{a-bi}/{a-bi}={a-bi}/{a^2+b^2}#


    I hope that this was helpful.

  • If complex numbers #z_1# and #z_2# in polar form are

    #{(z_1=r_1(costheta_1+isin theta_1)),(z_2=r_2(cos theta_2+i sin theta_2)):}#,

    then we can write in exponential form

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

    So, the quotient #z_1/z_2# can be written as

    #z_1/z_2={r_1e^{i theta_1}}/{r_2e^{i theta_2}}=r_1/r_2e^{i(theta_1-theta_2)}#

    #=r_1/r_2[cos(theta_1-theta_2)+isin(theta_1-theta_2)]#


    I hope that this was helpful.

  • Let #z_1 = a_1+b_1i# and #z_2=a_2+b_2i#. We want to find

    #q=z_1/z_2=(a_1+b_1i)/(a_2+b_2i)#

    Generally, we wish to write this in the form

    #q=A+Bi#

    Where #A# and #B# are real numbers. To do this, we must amplify the quotient by the conjugate of the denominator:

    #q=z_1/z_2 * bar(z_2)/(bar(z_2))=(a_1+b_1i)/(a_2+b_2i)*(a_2-b_2i)/(a_2-b_2i)=((a_1a_2+b_1b_2)+(b_1a_2-b_2a_1)i)/(a_2^2+b_2^2)#

    #q = (a_1a_2+b_1b_2)/(a_2^2+b_2^2) + (b_1a_2-b_2a_1)/(a_2^2+b_2^2) i#

Questions