# How do you add (5+4i)+(6+i) in trigonometric form?

Feb 19, 2018

$11 + 5 i$

#### Explanation:

Complex numbers in the form $a + b i$ can be represented as:

$z = r \left(\cos \theta + i \sin \theta\right)$

Where:

$\boldsymbol{r} = \sqrt{{a}^{2} + {b}^{2}}$

$\boldsymbol{\theta} = \arctan \left(\frac{b}{a}\right)$

We now put our complex numbers in this form:

Let:

${z}_{1} = 5 + 4 i$ and ${z}_{2} = 6 + i$

For ${z}_{1}$

$r = \sqrt{{\left(5\right)}^{2} + {\left(4\right)}^{2}} = \sqrt{41}$

$\theta = \arctan \left(\frac{4}{5}\right) = 38.66$

${z}_{1} = \sqrt{41} \left(\cos \left(38.66\right) + i \sin \left(38.66\right)\right)$

For ${z}_{2}$

$r = \sqrt{{\left(6\right)}^{2} + {\left(1\right)}^{1}} = \sqrt{37}$

$\theta = \arctan \left(\frac{1}{6}\right) = 9.46$

${z}_{2} = \sqrt{37} \left(\cos \left(9.46\right) + i \sin \left(9.46\right)\right)$

Addition is the same as for that of complex numbers in the form $a + b i$

$\therefore$

${a}_{1} + {b}_{1} i + {a}_{2} + {a}_{2} i = \left({a}_{1} + {a}_{1}\right) + \left({b}_{1} + {b}_{2}\right) i$

${z}_{1} + {z}_{2}$

sqrt(41)(cos(38.66)+sqrt(37)(cos(9.46)=11.00002714

isqrt(41)(sin(38.66)+isqrt(37)sin(9.46)=4.999773551i

Rounding to 2 .d.p.

$11.00 + 5.00 i = 11 + 5 i$

Notice the result is exactly the same as adding in rectangular form.

$\left(5 + 4 i\right) + \left(6 + i\right) = \left(5 + 6\right) + \left(6 + 1\right) i = 11 + 5 i$