Division of Complex Numbers
Key Questions
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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
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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.
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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.
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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#