Question

express the following complex number in the form r(cos theta + i sin theta):

(i) 1+i tan alpha

(ii) tan alpha - i​

Answer 1

Polar form of Z

Case 1st

1+ i tan alpha

we know that r= |z|

First of all we have to find value of |z|

$|z| = \sqrt{re {(z)}^{2} + im {(z)}^{2} }$

$|z| = \sqrt{ {(1)}^{2} + ( { \tan( \ \alpha ) }^{2}) }$

1+ tan²alpha = sec²alpha

$|z| = \sec( \alpha )$

Now we have to find arg(z)

As z lies in 1st quadrant so theta will

Arg(z) =

$\tan( \theta ) = | \frac{im(z)}{re(z)} |$

$\tan( \theta) = | \frac{ \tan( \alpha ) }{1} |$

tan theta = tan alpha

theta = alpha

r = sec alpha

theta = alpha

Polar form of Z will

Z = r (cos theta + I sin theta)

Z = sec alpha ( cos alpha + I sin alpha)

Case 2nd

$\tan( \alpha ) - i$

Here r = |z| = formula mentioned above

$|z| = \sqrt{ \tan( { \alpha }^{2} ) + ( {1)}^{2} }$

as we solved above..

$|z| = \sec( { \alpha })$

Arg(z) =

$\tan( \theta) = | \frac{ - 1}{ \tan( \alpha ) } |$

$\tan( \theta) = \frac{1}{ \tan( \alpha ) }$

$\tan( \theta) = \cot( \alpha )$

As we know that cot theta = 1/tan theta

if alpha <90 then

$\tan( \theta) = \tan(90 - \alpha )$

$\theta = \frac{\pi}{2} - \alpha$

As Z lies in 4th quadrant

$\theta = - (\pi - \frac{\pi}{2} + \alpha )$

$\theta = (\frac{\pi}{2} + \alpha )$

So,

Z = sec alpha [ cos( π/2 +alpha) + I sin (π/2 +alpha)]

cos ( 90+ alpha ) =- sin alpha

Sin ( 90+alpha ) = cos alpha

So,

Z = sec alpha (- sin alpha + i cos alpha)

If alpha >90, then

$\tan( \theta) = \tan(90 + \alpha )$

$(\theta) = \frac{\pi}{2} + \alpha$

as z lies in 4th quadrant

theta = -(π- π/2-alpha)

theta = - π/2-alpha

Polar form of Z will

Z = sec alpha [cos (π/2 - alpha) + i sin ( π/2- alpha)]

$\cos( \frac{\pi}{2} - \alpha ) = \sin( \alpha )$

$\sin( \frac{\pi}{2} - \alpha ) = \cos( \alpha )$

So ,

Z = sec alpha ( sin alpha + I cos alpha)

Hope it helps you

Answer 2

Step-by-step explanation:

Case 1st

1+ i tan alpha

we know that r= |z|

First of all we have to find value of |z|

|z| = \sqrt{re {(z)}^{2} + im {(z)}^{2} }∣z∣=

re(z)

2

+im(z)

2

|z| = \sqrt{ {(1)}^{2} + ( { \tan( \ \alpha ) }^{2}) }∣z∣=

(1)

2

+(tan( α)

2

)

1+ tan²alpha = sec²alpha

|z| = \sec( \alpha )∣z∣=sec(α)

Now we have to find arg(z)

As z lies in 1st quadrant so theta will

Arg(z) =

\tan( \theta ) = | \frac{im(z)}{re(z)} |tan(θ)=∣

re(z)

im(z)

∣

\tan( \theta) = | \frac{ \tan( \alpha ) }{1} |tan(θ)=∣

1

tan(α)

∣

tan theta = tan alpha

theta = alpha

r = sec alpha

theta = alpha

Polar form of Z will

Z = r (cos theta + I sin theta)

Z = sec alpha ( cos alpha + I sin alpha)

Case 2nd

\tan( \alpha ) - itan(α)−i

Here r = |z| = formula mentioned above

|z| = \sqrt{ \tan( { \alpha }^{2} ) + ( {1)}^{2} }∣z∣=

tan(α

2

)+(1)

2

as we solved above..

|z| = \sec( { \alpha })∣z∣=sec(α)

Arg(z) =

\tan( \theta) = | \frac{ - 1}{ \tan( \alpha ) } |tan(θ)=∣

tan(α)

−1

∣

\tan( \theta) = \frac{1}{ \tan( \alpha ) }tan(θ)=

tan(α)

1

\tan( \theta) = \cot( \alpha )tan(θ)=cot(α)

As we know that cot theta = 1/tan theta

if alpha <90 then

\tan( \theta) = \tan(90 - \alpha )tan(θ)=tan(90−α)

\theta = \frac{\pi}{2} - \alphaθ=

2

π

−α

As Z lies in 4th quadrant

\theta = - (\pi - \frac{\pi}{2} + \alpha )θ=−(π−

2

π

+α)

\theta = (\frac{\pi}{2} + \alpha )θ=(

2

π

+α)

So,

Z = sec alpha [ cos( π/2 +alpha) + I sin (π/2 +alpha)]

cos ( 90+ alpha ) =- sin alpha

Sin ( 90+alpha ) = cos alpha

So,

Z = sec alpha (- sin alpha + i cos alpha)

If alpha >90, then

\tan( \theta) = \tan(90 + \alpha )tan(θ)=tan(90+α)

(\theta) = \frac{\pi}{2} + \alpha(θ)=

2

π

+α

as z lies in 4th quadrant

theta = -(π- π/2-alpha)

theta = - π/2-alpha

Polar form of Z will

Z = sec alpha [cos (π/2 - alpha) + i sin ( π/2- alpha)]

\cos( \frac{\pi}{2} - \alpha ) = \sin( \alpha )cos(

2

π

−α)=sin(α)

\sin( \frac{\pi}{2} - \alpha ) = \cos( \alpha )sin(

2

π

−α)=cos(α)

So ,

Z = sec alpha ( sin alpha + I cos alpha)

Hope it helps you