How do you convert #1 + 4i# to polar form?

1 Answer
Alan N.
May 15, 2016

#~=sqrt(17)(cos1.33 + isin1.33)#

Explanation:

If #z = a +bi#

Then #z# may be expressed in polar form as:

#z = r(cos theta + i sin theta)#
Where: #r = sqrt(a^2 + b^2)# and #theta = arctan(b/a)#

In this example: #z=1+4i#
Hence: #a=1# and #b=4#

Therefore: #r = sqrt(1^2+4^2)# =#sqrt(17)#
And: #theta = arctan(4/1)# #~= 1.33#

Hence: #z ~= sqrt(17)(cos 1.33 + i sin 1.33)#

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