Question

convert the complex number (1+7i)/(3-4i) in polar form

Answer 1

GIVEN :

The complex number is $\frac{1+7i}{3-4i}$

TO FIND :

The polar form for the given complex number.

SOLUTION :

Given that the  complex number is $\frac{1+7i}{3-4i}$

Now rationalising the given complex number as below,

$\frac{1+7i}{3-4i}$

$=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}$

$=\frac{(1+7i)(3+4i)}{(3-4i)(3+4i)}$

By using the Distributive property:

(x+y)(a+b)=(x+y)a+(x+y)b

$=\frac{(1+7i)(3)+(1+7i)(4i)}{(3-4i)(3+4i)}$

By using the Distributive property:

a(x+y)=ax+ay

By using the Algebraic identity:

$(a-b)(a+b)=a^2-b^2$

$=\frac{1(3)+1(4i)+7i(3)+7i(4i)}{3^2-(4i)^2}$

$=\frac{3+4i+21i+28i^2}{9-16i^2}$

$=\frac{3+25i+28(-1)}{9-16(-1)}$ (since $i^2=-1$)

$=\frac{3+25i-28}{9+16}$

$=\frac{25i-25}{25}$

$=-1+i$

∴ z=-1+i

Let the polar form be $z=r(cos\theta+isin\theta)$

Now we have that,

$-1+i=r(cos\theta+isin\theta)$

$-1+i=rcos\theta+irsin\theta$

Equating the real and imaginary parts

$-1=rcos\theta$          ,                         $1=rsin\theta$

Squaring on both sides,        Squaring on both sides

$-1^2=(rcos\theta)^2$     ,                    $1^2=(rsin\theta)^2$

$1=r^2cos^2\theta\hfill (1)$     ,      $1=r^2sin^2\theta\hfill (2)$

Adding the equations (1) and (2)

$1+1=r^2cos^2\theta+r^2sin^2\theta$

$2=r^2(cos^2\theta+sin^2\theta)$

By using the trignometric identity

$cos^2x+sin^2x=1$

$2=r^2(1)$

$2=r^2$

$r^2=2$

∴ $r=\sqrt{2}$

Now we have to find the argument of z:

$-1+i=r(cos\theta+isin\theta)$

Substitute the value of r,

$-1+i=\sqrt{2}(cos\theta+isin\theta)$

$-1+i=\sqrt{2}cos\theta+\sqrt{2}isin\theta$

Equating the real and imaginary parts

$-1=\sqrt{2}cos\theta$     ,      $1=\sqrt{2}sin\theta$

∴ $cos\theta=-\frac{1}{\sqrt{2}}$  ,   $sin\theta=\frac{1}{\sqrt{2}}$

Since $sin\theta$ is positive  and $cos\theta$ is negative,

∴ $\theta$ lies in 2nd quadrant.

$Argument=180^{\circ}-45^{\circ}$

$=135^{\circ}$

$=135^{\circ}\times \frac{\pi}{180^{\circ}}$

$=\frac{3\pi}{4}$

∴   $arg(z)=\frac{3\pi}{4}$

We have that $r=\sqrt{2}$ and $\theta=\frac{3\pi}{4}$

The polar form of z is

$z=r(cos\theta+isin\theta)$

$z=\sqrt{2}(cos\frac{3\pi}{4}+isin\frac{3\pi}{4})$

∴ The polar form of z is $\sqrt{2}(cos\frac{3\pi}{4}+isin\frac{3\pi}{4})$