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Convert to Complex Number Cartesian Form
from Polar Form, Exponential Form

On the  complex numbers polar form  page, we see examples of converting from complex number cartesian form to complex number polar form.

CARTESIAN FORM:      z = a + bi

POLAR FORM:            z = r(cosθ + isinθ)

Converting the other way from polar form to complex number cartesian form is also possible.

To see this in action, we can look at examples (1.1) and (1.2) from the complex numbers polar form page.


Polar Form:   2√2(cos(\bf{\frac{\pi}{4}}) + isin(\bf{\frac{\pi}{4}}))

To convert to Cartesian Form.
We can multiply out as it sits, using the exact values for the cos and sin.

z  =  (2√2 × cos(\bf{\frac{\pi}{4}}))  +  i(2√2 × sin(\bf{\frac{\pi}{4}}))

z  =  (2√2 × \bf{\frac{1}{\sqrt{2}}})  +  i(2√2 × \bf{\frac{1}{\sqrt{2}}})

z  =  2 + i2


Polar Form:   √17(cos(1.816) + isin(1.816))

Now as the radians here are rounded to 3 decimal places, the initial calculations will be slightly out.

But the numbers obtained will be close enough to round both down and up to the given original values in this example from the polar form page.

For Cartesian Form:

z  =  (√17 × cos(1.816))  +  i(√17 × sin(1.816))

z  =  –1.0009…  +  i3.9997…

Which can be rounded to.

z  =  –1 + i4

Complex Number Cartesian Form,
Complex Number Exponential Form

It should also be mentioned that a complex number can also be expressed in “Exponential Form”.

We can also convert from and to this exponential form.

If we observe Euler’s Formula.

eiθ  =  cosθ + isinθ

A complex number can written as   z = reiθ.

Where  r  is once again the modulus of the complex number.

With  θ  again being the argument, specifically in radians.



Write in exponential form   z = 1 + i√3.


r  =  \bf{\sqrt{1^2 \space + \space (\sqrt{3})^2}}  =  2

a > 0 , b > 0   =>    θ = tan-1(\bf{\frac{\sqrt{3}}{1}})  =  \bf{\frac{\pi}{3}}

In exponential form:    z = 2ei\bf{\frac{\pi}{3}}


Write in exponential form

z  =  2√2(cos(\bf{\frac{\pi}{4}}) + isin(\bf{\frac{\pi}{4}})).


In exponential form:    z2√2ei\bf{\frac{\pi}{4}}


Write in cartesian form   z  =  4ei\bf{\frac{\pi}{6}}.


Put the complex number into polar form first.

z  =  4(cos(\bf{\frac{\pi}{6}}) + isin(\bf{\frac{\pi}{6}}))

Then carry out the calculation.

z  =  (4 × cos(\bf{\frac{\pi}{6}}))  +  i(4 × sin(\bf{\frac{\pi}{6}}))

z  =  (4 x \bf{\frac{\sqrt{3}}{2}})  +  i(4 × \bf{\frac{1}{2}})

z  =  2√3 + i2

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