What is the trigonometric form of # (3+5i) #?

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Answer 1 · 2015-12-27T17:14:19

#3+5i = sqrt(34)(cos(tan^(-1)(5/3)) + isin(tan^(-1)(5/3)))#

#~~ sqrt(34)(cos(59.04^@) + isin(59.04^@))#

Any complex number #z = a+bi# has a trigonometric form

#z = r(cos(theta) + isin(theta))#

where #r = |z| = sqrt(a^2 + b^2)# and #theta = tan^(-1)(b/a)#

For the given complex number, we have #a = 3# and #b = 5#. Thus

#r = sqrt(3^2 + 5^2) = sqrt(9+25) = sqrt(34)#

and

#theta = tan^(-1)(5/3) ~~ 59.04^@#

So we have the trigonometric form

#3+5i = sqrt(34)(cos(tan^(-1)(5/3)) + isin(tan^(-1)(5/3)))#

#~~ sqrt(34)(cos(59.04^@) + isin(59.04^@))#