Question

Chapter Name :- Complex Numbers


If $\pink{\bf re^{i\theta} = \dfrac{3+2i}{2+3i} + \dfrac{1+5i}{1-2i} }$ then find out the value of $\pink{\bf r}$ and $\pink{\bf\theta}$.




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Answer 1

Step-by-step explanation:

First, let simplify the operation.

$\frac{3 + 2i}{2 + 3i} + \frac{1 + 5i}{1 - 2i}$

$\frac{(3 + 2i)(1 - 2i)}{(2 + 3i)(1 - 2i)} + \frac{(1 + 5i)(2 + 3i)}{(1 - 2i)(2 + 3i)} = \frac{3 - 4i + 4 + 2 + 13i - 15}{2 - i + 6}$

Which then gets us

$\frac{ - 6 + 9i}{ 8 - i}$

Which then gets us

$\frac{ - 6 + 9i(8 + i)}{8 - i(8 + i)} = \frac{ - 48 - 6i + 72i - 9}{65} = \frac{ - 57 + 66i}{65}$

So our equation is

$- \frac{57}{65} + \frac{66i}{65}$

Now know that re^i(theta) is the same as

The trig form of a complex number, which is

we only interested in the radius of the angle with respect to the origin and the angle of the rotation of the complex number.

First, using Pythagoras find the r.

$r = \sqrt{ (\frac{ - 57}{65} {}^{} ) {}^{2} + (\frac{66}{65} ) {}^{2} }$

We get

$\frac{3 \sqrt{5} }{5}$

So our r value is

$\frac{3 \sqrt{5} }{5}$

Now we must find our angle of rotation. Apply the tangent rule of finding rotation using complex number.

If we have a complex number a+bi, the angle of rotation is

$\tan(x) = \frac{b}{a}$

Here our b value is 66/65 and a is -57/65.

So we subsitue

$\tan(x) = - \frac{66}{57}$

Remeber same denomiator when dividng cancel out.

We then get x=130.82 degrees or 2.28 rads.

Since our real number,a, of the complex number is negative, if we graph this complex number, it will be in the second quadrant on the Argand plane. This is our answer because the degree 130 falls in the second quadrant where degree are between 90 and 180 degree or pi/2 and pi rads. 2.28 also falls between 1.57 rads and 3.14 rads.

So our value of Theta is

$130.82$

degrees or

$2.28$

rads.

So we have

$( \frac{3 \sqrt{5} }{5} )e {}^{2.28i}$

Answer 2

The value of the modulus and argument are 16.64 and 61.28degrees respectively

Modulus and argument of complex numbers

The formula for calculating the modulus and argument of a complex number $z=x+iy$ is given as:

$|z| =\sqrt{x^2+y^2}\\ \theta =tan^{-1}\frac{y}{x}$

Given the complex expression

$\frac{3+2i}{2+3i} +\frac{1+5i}{1-2i}$

Simplify and write in the form x + iy as shown:

$=\frac{(3+2i)(1+5i)}{(2+3i)(1-2i)} \\ =\frac{3+17i-10}{3-i+6}\\ =\frac{-7+17i}{-+i}$

Rationalize the result

$=\frac{-7+17i}{9-i} \times \frac{9+i}{9+i} \\ =\frac{-80-146i}{10} \\ =-8-14.6i$

Find the modulus

$|z| =\sqrt{8^2+14.6^2}\\ |z| =\sqrt{277.16}\\ |z|=16.64$

Get the argument:

$\theta = tan^{-1}\frac{14.6}{8} \\ \theta = 61.28^0$

Hence the value of the modulus and argument are 16.64 and 61.28degrees respectively

Learn more on complex numbers here: https://brainly.com/question/10662770