Convert the following in the polar form:

We have,
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= ![]()
Now, by multiplying numerator and denominator by 3 + 4i we get:
= ![]()
= ![]()
= ![]()
= ![]()
= - 1 + i
Let us assume
and ![]()
Now by squaring and adding the given values, we get:
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We know that, ![]()
∴ r2 = 2
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As conventionally r > 0, thus ![]()
Also, ![]()
As
lies in the 2nd quadrant
∴ ![]()
Hence, ![]()
= ![]()
= ![]()
This is the required polar form of the given equation