How do you multiply # (9-i)(7-3i) # in trigonometric form?

1 Answer
Shwetank Mauria · Anthony X
Feb 12, 2016

#2sqrt(1189)(cos (alpha+beta)+isin(alpha+beta))#, where #alpha=tan^(-1)(-1/9)# and #beta=tan^(-1)(-3/7)#

Explanation:

A complex number of form #a+bi# can be written as #r(cos theta +i sin theta)#, where #r=sqrt(a^2+b^2)# and #theta=tan^(-1)(b/a)#.

Using this #(9-i)=sqrt82(cos alpha+i sin alpha)#, and #alpha=tan^(-1)(-1/9)#

Similarly, #(7-3i)=sqrt58(cos beta+i sin beta)#, and #beta=tan^(-1)(-3/7)#

Multiplication of #(sqrt82(cos alpha+i sin alpha)# and #sqrt58(cos beta+i sin beta)# is given by

#2sqrt(1189)(cos (alpha+beta)+isin(alpha+beta))#, where #alpha=tan^(-1)(-1/9)# and #beta=tan^(-1)(-3/7)#, as #sqrt(82*58)=2sqrt(1189)#.

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