Complex arithmetic with conjugates Show that the conjugate of the sum (product, or quotient) of two complex numbers, z1 and z2, is the same as the sum (product, or quotient) of their conjugates.

Question

Complex arithmetic with conjugates Show that the conjugate  of the sum (product, or quotient) of two complex numbers, z1 and z2, is the same as the sum (product, or quotient) of their conjugates.

Wendy Boykin · Accepted Answer

Step 1  Let two complex number are  z1=a+ib  z2=c+id  Step 2  Sum of these two complex numbers  z1+z2=(a+ib)+(c+id)  =(a+c)+i(b+c)  Product of two complex numbers  z1×z2=(a+ib)(c+id)  =(ac+iad+ibc+i2bd)  =(ac+iad+ibc−bd)  =(ac−bd)+i(ad+bc)  Step 3  The quotient of two complex numbers  z1z2=a+ibc+id  =a+ibc+id×c−idc−id  =ac−iad+ibc−i2bdc2−i2d2  =ac+i(bc−ad)+bdc2+d2  =ac+bd+i(bc−ad)c2+d2  =ac+bdc2+d2+ibc−adc2+d2  Hence,  The conjugate of the sum (product, or quotient) of two complex numbers, z1 and z2, is the same as the sum (product, or quotient) of their conjugates.

Kirsten Davis · Accepted Answer

Step 1  Let z1=a+ib and z2=c+id be two complex numbers.  Then we get,  Conjugate of z1 is K=a−ib  Conjugate of z2 is =c−id  Sum of the conjugates of z1 and z2=(a−ib)+(c−id)=(a+c)−i(b+d)  Step 2  Moreover we see that,  z1+z2=(a+ib)+(c+id)=(a+c)+i(b+d)  Therefore the conjugate of the sum of z1 and z2 is:  Conjugate of (z1+z2) = Conjugate of [(a+ib)+(c+id)]= Conjugate of [(a+c)+i(b+d)]=(a+c)−i(b+d)   Therefore,  Sum of the conjugates of z1 and z2 = Conjugate of sum of z1 and z2  Hence the proof.

Vasquez · Accepted Answer

Step 1Consider two complex numbers a+bi and c+di where a,b,c,d are real numbersAdd both complex numbers(a+bi)+(c+di)=a+bi+c+di=a+c+bi+di=(a+c)+i(b+d)Step 2Subtract both complex numbers(a+bi)-(c+di)=a+bi-c-di=a-c+bi-di=(a-c)+i(b-d)Hence &amp;ldquo;it makes sense&amp;rdquo; that it&amp;#39;s just combining like terms that is real terms and imaginary terms.

Complex arithmetic with conjugates Show that the conjugate of the sum

Answered question

Answer & Explanation

New Questions in Precalculus