CALCULUS ASSIGNMENT COMPLEX NUMBERS

RANA MUHAMMED AHMED BILAL EE-160

Department of Electrical Engineering

JANUARY 27, 2016

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

1

Assignment Question # 1

𝑨𝒑𝒑𝒍𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝑫𝒆 𝑴𝒐𝒊𝒗𝒆𝒓′𝒔𝑻𝒉𝒆𝒐𝒓𝒆𝒎

(𝒄𝒐𝒔 ∝ +𝒊 𝒔𝒊𝒏 𝜶)( 𝒄𝒐𝒔 𝜷 + 𝒊 𝒔𝒊𝒏 𝜷)

(𝒄𝒐𝒔 𝜸 + 𝒊 𝒔𝒊𝒏 𝜸)( 𝒄𝒐𝒔 𝝋 + 𝒊 𝒔𝒊𝒏 𝝋)= 𝒄𝒊𝒔(𝜶 + 𝜷 − 𝜸 − 𝝋) ⟶ 𝑨

𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏:

Consider: (𝒄𝒐𝒔 ∝ +𝒊 𝒔𝒊𝒏 𝜶)( 𝒄𝒐𝒔 𝜷 + 𝒊 𝒔𝒊𝒏 𝜷)

upon multiplying:

= cos 𝛼 cos 𝛽 + cos 𝛼 𝑖 sin 𝛽 + cos 𝛽 𝑖 sin 𝛼 + sin 𝛼 sin 𝛽 𝑖2

= cos 𝛼 cos 𝛽 + 𝑖 cos 𝛼 sin 𝛽 + icos 𝛽 sin 𝛼 − sin 𝛼 sin 𝛽

= (𝒄𝒐𝒔 𝜶 𝒄𝒐𝒔 𝜷 − 𝒔𝒊𝒏 𝜶 𝒔𝒊𝒏 𝜷) + 𝒊(𝒄𝒐𝒔 𝜶 𝒔𝒊𝒏 𝜷 + 𝒄𝒐𝒔 𝜷 𝒔𝒊𝒏 𝜶)

Using formulae’s

(𝒄𝒐𝒔 ∝ +𝒊 𝒔𝒊𝒏 𝜶)( 𝒄𝒐𝒔 𝜷 + 𝒊 𝒔𝒊𝒏 𝜷) = 𝒄𝒐𝒔(𝜶 + 𝜷) + 𝒊 𝒔𝒊𝒏(𝜶 + 𝜷) → 1

Now; consider: (𝒄𝒐𝒔 𝜸 + 𝒊 𝒔𝒊𝒏 𝜸)( 𝒄𝒐𝒔 𝝋 + 𝒊 𝒔𝒊𝒏 𝝋) Upon Multiplying:

=cos 𝛾 cos 𝜑 + cos 𝛾 𝑖 sin 𝜑 + 𝑖 sin 𝛾 cos 𝜑 + 𝑖2 sin 𝛾 sin 𝜑

= cos 𝛾 cos 𝜑 + 𝑖 cos 𝛾 sin 𝜑 + 𝑖 sin 𝛾 cos 𝜑 − sin 𝛾 sin 𝜑

= (𝒄𝒐𝒔 𝜸 𝒄𝒐𝒔 𝝋 − 𝒔𝒊𝒏 𝜸 𝒔𝒊𝒏 𝝋) + 𝒊(𝒄𝒐𝒔 𝜸 𝒔𝒊𝒏 𝝋 + 𝒔𝒊𝒏 𝜸 𝒄𝒐𝒔 𝝋)

Using formulae’s

(𝒄𝒐𝒔 𝜸 + 𝒊 𝒔𝒊𝒏 𝜸)( 𝒄𝒐𝒔 𝝋 + 𝒊 𝒔𝒊𝒏 𝝋) = 𝒄𝒐𝒔(𝜸 + 𝝋) + 𝒊 𝒔𝒊𝒏(𝜸 + 𝝋) → 2

𝒑𝒖𝒕𝒕𝒊𝒏𝒈 𝒗𝒂𝒍𝒖𝒆𝒔 𝒇𝒓𝒐𝒎 𝟏 & 𝟐 𝒊𝒏𝒕𝒐 𝒆𝒒 ′𝑨′

→ (cos ∝ +𝑖 sin 𝛼)( cos 𝛽 + 𝑖 sin 𝛽)

(cos 𝛾 + 𝑖 sin 𝛾)( cos 𝜑 + 𝑖 sin 𝜑)=

cos(𝛼 + 𝛽) + 𝑖 sin(𝛼 + 𝛽)

cos(𝛾 + 𝜑) + 𝑖 sin(𝛾 + 𝜑)

Using De Moiver’s theorem:

{cos(𝛼 + 𝛽) + 𝑖 sin(𝛼 + 𝛽)} {cos(𝛾 + 𝜑) + 𝑖 sin(𝛾 + 𝜑)}-1

{𝒄𝒐𝒔(𝜶 + 𝜷) + 𝒊 𝒔𝒊𝒏(𝜶 + 𝜷)}{𝒄𝒐𝒔(−𝜸 − 𝝋) + 𝒊 𝒔𝒊𝒏(−𝜸 − 𝝋)}

Let u = (𝛼 + 𝛽)

And v =(−𝛾 − 𝜑)

Then the equation will transform as follows

(𝒄𝒐𝒔 𝒖 + 𝒊 𝒔𝒊𝒏 𝒖)(𝒄𝒐𝒔 𝒗 + 𝒊 𝒔𝒊𝒏 𝒗)

= cos 𝑢 cos 𝑣 + 𝑖 cos 𝑢 sin 𝑣 + 𝑖 sin 𝑢 cos 𝑣 + 𝑖2 sin 𝑢 sin 𝑣

= cos 𝑢 cos 𝑣 − sin 𝑢 sin 𝑣 + 𝑖(sin 𝑢 cos 𝑣 + cos 𝑢 sin 𝑣)

using formulae’s;

= 𝒄𝒐𝒔(𝒖 + 𝒗) + 𝒊 𝒔𝒊𝒏(𝒖 + 𝒗)

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

2

putting values of u and v

= cos(𝛼 + 𝛽 − 𝛾 − 𝜑) + 𝑖 sin(𝛼 + 𝛽 − 𝛾 − 𝜑)

= 𝑪𝒊𝒔(𝜶 + 𝜷 − 𝜸 − 𝝋)

↔ 𝐏𝐫𝐨𝐯𝐞𝐝

Assignment Question # 2

𝒊𝒏𝒗𝒆𝒓𝒔𝒆 𝒉𝒚𝒑𝒆𝒓𝒃𝒐𝒍𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏:

I. 𝒄𝒐𝒔𝒉−𝟏 𝒛 = 𝒍𝒏(𝒛 + √𝒛𝟐 − 𝟏)

Solution:

⇒ 𝐿. 𝐻. 𝑆

Let 𝑥 = cosh−1 𝑧

𝑧 = cosh 𝑥

We know that

cosh2 𝑥 − sinh2 𝑥 = 1

𝑧2 − sinh2 𝑥 = 1

sinh 𝑥 = √𝑧2 − 1 ⟶ 1

We also know that

𝑒𝑥 = sinh 𝑥 + cosh 𝑥

𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑟𝑜𝑚 1

𝑒𝑥 = √𝑧2 − 1 + 𝑧

Re arranging

𝑒𝑥 = 𝑧 + √𝑧2 − 1

ln 𝑒𝑥 = ln (𝑧 + √𝑧2 − 1 )

𝑥 ln 𝑒 = ln(𝑧 + √𝑧2 − 1)

𝑥 = ln(𝑧 + √𝑧2 − 1) ∵ ln 𝑒 = 1

csch−1 𝑧 = ln (𝑧 + √𝑧2 − 1) ∎ ∵ 𝑥 = csch−1 𝑧

II. 𝒄𝒐𝒕𝒉−𝟏 𝒛 =𝟏

𝟐𝒍𝒏 (

𝒁+𝟏

𝒁−𝟏 )

Solution:

𝐿𝑒𝑡

𝑥 = coth−1 𝑧

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

3

𝑧 = coth 𝑥

𝑧 =cosh 𝑥

sinh 𝑥

𝑧 =

𝑒𝑥 + 𝑒−𝑥

2𝑒𝑥 − 𝑒−𝑥

2

∵ cosh 𝑥 =𝑒𝑥 + 𝑒−𝑥

2; sinh 𝑥 =

𝑒𝑥 − 𝑒−𝑥

2

z =𝑒𝑥 + 𝑒−𝑥

𝑒𝑥 − 𝑒−𝑥

⇒ 𝑧 =𝑒𝑥 + 𝑒−𝑥

𝑒𝑥 − 𝑒−𝑥×

𝑒𝑥

𝑒𝑥

𝑧 =𝑒2𝑥 + 𝑒−𝑥+𝑥

𝑒2𝑥 − 𝑒−𝑥+𝑥

𝑧 =𝑒2𝑥 + 𝑒0

𝑒2𝑥 − 𝑒0⟹ 𝑧 =

𝑒2𝑥 + 1

𝑒2𝑥 − 1

𝑧(𝑒2𝑥 − 1) = 𝑒2𝑥 + 1

𝑧𝑒2𝑥 − 𝑧 = 𝑒2𝑥 + 1

𝑧𝑒2𝑥 − 𝑒2𝑥 = 𝑧 + 1

𝑒2𝑥(𝑧 − 1) = 𝑧 + 1

𝑒2𝑥 = (𝑧 + 1)

(𝑧 − 1)

ln 𝑒2𝑥 = ln(𝑧 + 1)

(𝑧 − 1)

2𝑥 ln 𝑒 = ln(𝑧 + 1)

(𝑧 − 1) ∵ ln 𝑒 = 1

2𝑥 = ln(𝑧 + 1)

(𝑧 − 1)

𝑥 =1

2ln

(𝑧 + 1)

(𝑧 − 1)

coth−1 𝑧 =1

2ln

(𝑧 + 1)

(𝑧 − 1)∎ ∵ 𝑥 = coth−1 𝑧

III. 𝒔𝒆𝒄𝒉−𝟏 𝒛 = 𝒍𝒏 (𝟏+√𝟏−𝒛𝟐

𝒛)

Solution:

𝐿𝑒𝑡

𝑥 = sech−1 𝑧

𝑧 = sech 𝑥

Using

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

4

1 − tanh2 𝑥 = sech2 𝑥

1 − sech2 𝑥 = tanh2 𝑥

1 − 𝑧2 = tanh2 𝑥

tanh 𝑥 = √1 − 𝑧2 ⟶ 1

𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡

𝑒𝑥 = cosh 𝑥 + sinh 𝑥

𝑒𝑥 =1

sech 𝑥+

sinh 𝑥

cosh 𝑥× cosh 𝑥 ∵ cosh 𝑥 =

1

sech 𝑥

𝑒𝑥 =1

sech 𝑥+ tanh 𝑥. cosh 𝑥 ∵

sinh 𝑥

cosh 𝑥= tanh 𝑥

𝑒𝑥 =1

sech 𝑥+

tanh 𝑥

sech 𝑥

𝑒𝑥 =1 + tanh 𝑥

sech 𝑥

𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑟𝑜𝑚 𝑒𝑞 1

𝑒𝑥 =1 + √1 − 𝑧2

𝑧

𝑇𝑎𝑘𝑖𝑛𝑔 𝑙𝑛 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒

ln 𝑒𝑥 = ln (1 + √1 − 𝑧2

𝑧)

𝑥 ln 𝑒 = ln (1 + √1 − 𝑧2

𝑧)

𝑥 = ln (1 + √1 − 𝑧2

𝑧)

sech−1 𝑥 = ln (1 + √1 − 𝑧2

𝑧) ∎

IV. 𝒄𝒔𝒄𝒉−𝟏 𝒛 = 𝒍𝒏 (𝟏+√𝟏+𝒛𝟐

𝒛)

Solution:

𝑳𝒆𝒕

𝑥 = csch−1 𝑧

csch 𝑥 = 𝑧

1 + csch2 𝑥 = coth2 𝑥

1 + 𝑧2 = coth2 𝑥

coth 𝑥 = √1 + 𝑧2 ⟶ 1

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

5

𝑢𝑠𝑖𝑛𝑔

𝑒𝑥 = cosh 𝑥 + sinh 𝑥

𝑒𝑥 = sinh 𝑥 ×cosh 𝑥

sinh 𝑥+ sinh 𝑥

𝑒𝑥 = sinh 𝑥 . coth 𝑥 + sinh 𝑥

𝑒𝑥 =1

csch 𝑥. coth 𝑥 +

1

csch 𝑥

𝑒𝑥 =coth 𝑥

csch 𝑥+

1

csch 𝑥

𝑒𝑥 =1 + coth 𝑥

csch 𝑥

𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑟𝑜𝑚 1

𝑒𝑥 =1 + √1 + 𝑧2

𝑧

𝑇𝑎𝑘𝑖𝑛𝑔 𝑙𝑛 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒

ln 𝑒𝑥 = ln (1 + √1 + 𝑧2

𝑧)

𝑥 ln 𝑒 = ln (1 + √1 + 𝑧2

𝑧)

𝑥 = ln (1 + √1 + 𝑧2

𝑧)

sech−1 𝑥 = ln (1 + √1 + 𝑧2

𝑧) ∎

Assignment Question # 3

𝒊𝒏𝒗𝒆𝒓𝒔𝒆 𝒕𝒓𝒊𝒈𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏:

I. 𝐜𝐨𝐬−𝟏 𝒛 =𝟏

𝒊𝐥𝐧(𝒛 + √𝒛𝟐 − 𝟏)

Solution:

𝑙𝑒𝑡

𝑥 = cos−1 𝑧

𝑧 = cos 𝑥 ⟶ 1

𝑢𝑠𝑖𝑛𝑔

sin2 𝑥 +cos2 𝑥 = 1

sin2 𝑥 = 1 − cos2 𝑥

sin 𝑥 = √1 − cos2 𝑥

𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑟𝑜𝑚 𝑒𝑞 1

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

6

sin 𝑥 = √1 − 𝑧2

𝑢𝑠𝑖𝑛𝑔

𝑒𝑖𝑥 = cos 𝑥 + 𝑖 sin 𝑥

𝑒𝑖𝑥 = 𝑧 + 𝑖√1 − 𝑧2

𝑒𝑖𝑥 = 𝑧 + √−1√1 − 𝑧2 ∵ 𝑖 = √−1

𝑒𝑖𝑥 = 𝑧 + √𝑧2 − 1

𝑡𝑎𝑘𝑖𝑛𝑔 𝑙𝑛 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒

ln 𝑒𝑖𝑥 = ln (𝑧 + √𝑧2 − 1)

𝑖𝑥 ln 𝑒 = ln (𝑧 + √𝑧2 − 1)

𝑥 =1

𝑖ln (𝑧 + √𝑧2 − 1) ∵ ln 𝑒 = 1

cos−1 𝑧 =1

𝑖ln (𝑧 + √𝑧2 − 1) ∎

II. 𝒕𝒂𝒏−𝟏 𝒛 =𝟏

𝟐𝒊𝒍𝒏 (

𝟏+𝒊𝒛

𝟏−𝒊𝒛)

Solution:

𝑙𝑒𝑡

𝑥 = 𝑡𝑎𝑛−1 𝑧

tan 𝑥 = 𝑧

𝑧 =sin 𝑥

cos 𝑥

𝑧 =

𝑒𝑖𝑥 − 𝑒−𝑖𝑥

2𝑖𝑒𝑖𝑥 + 𝑒−𝑖𝑥

2

∵ 𝑆𝑖𝑛 𝑥 =𝑒𝑖𝑥 − 𝑒−𝑖𝑥

2𝑖; 𝐶𝑜𝑠 𝑥 =

𝑒𝑖𝑥 + 𝑒−𝑖𝑥

2

𝑧 =𝑒𝑖𝑥 − 𝑒−𝑖𝑥

𝑖(𝑒𝑖𝑥 + 𝑒−𝑖𝑥)

𝑧 =𝑒𝑖𝑥 − 𝑒−𝑖𝑥

𝑖(𝑒𝑖𝑥 + 𝑒−𝑖𝑥)×

𝑒𝑖𝑥

𝑒𝑖𝑥

𝑧 =𝑒𝑖𝑥+𝑖𝑥 − 𝑒−𝑖𝑥+𝑖𝑥

𝑖(𝑒𝑖𝑥+𝑖𝑥 + 𝑒−𝑖𝑥+𝑖𝑥)

𝑧 =𝑒2𝑖𝑥 − 𝑒0

𝑖(𝑒2𝑖𝑥 + 𝑒0)

𝑧 =𝑒2𝑖𝑥 − 1

𝑖(𝑒2𝑖𝑥 + 1) ∵ 𝑒0=1

𝑧𝑖(𝑒2𝑖𝑥 + 1) = 𝑒2𝑖𝑥 − 1

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

7

𝑧𝑒2𝑖𝑥 + 𝑧𝑖 = 𝑒2𝑖𝑥 − 1

𝑧𝑖𝑒2𝑖𝑥 − 𝑒2𝑖𝑥 = −𝑖𝑧 − 1

𝑒2𝑖𝑥(𝑧𝑖 − 1) = −(𝑖𝑧 + 1)

−𝑒2𝑖𝑥(𝑧𝑖 − 1) = (𝑖𝑧 + 1)

𝑒2𝑖𝑥(1 − 𝑖𝑧) = (1 + 𝑖𝑧)

𝑒2𝑖𝑥 =(1 + 𝑖𝑧)

(1 − 𝑖𝑧)

𝑡𝑎𝑘𝑖𝑛𝑔 ln 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒

ln 𝑒2𝑖𝑥 = ln(1 + 𝑖𝑧)

(1 − 𝑖𝑧)

2𝑖𝑥𝑙𝑛 𝑒 = ln(1 + 𝑖𝑧)

(1 − 𝑖𝑧)

𝑥 ln 𝑒 =1

2𝑖ln

(1 + 𝑖𝑧)

(1 − 𝑖𝑧)

tan−1 𝑧 =1

2𝑖ln

(1 + 𝑖𝑧)

(1 − 𝑖𝑧)∎ ∵ ln 𝑒 = 1

III. 𝐜𝐨𝐭−𝟏 𝐳 =𝟏

𝟐𝐢𝐥𝐧

( 𝐢+𝐳 )

( 𝐳−𝐢 )

Solution

𝑙𝑒𝑡

𝑥 = cot−1 𝑧

⇒ 𝑧 = cot 𝑥

𝑧 =cos 𝑥

sin 𝑥

𝑧 =

𝑒𝑖𝑥 + 𝑒−𝑖𝑥

2𝑒𝑖𝑥 − 𝑒−𝑖𝑥

2𝑖

∵ 𝑆𝑖𝑛 𝑥 =𝑒𝑖𝑥 − 𝑒−𝑖𝑥

2𝑖; 𝐶𝑜𝑠 𝑥 =

𝑒𝑖𝑥 + 𝑒−𝑖𝑥

2

𝑧 = (𝑒𝑖𝑥 + 𝑒−𝑖𝑥)𝑖

𝑒𝑖𝑥 − 𝑒−𝑖𝑥

𝑧 =(𝑒𝑖𝑥 + 𝑒−𝑖𝑥)𝑖

𝑒𝑖𝑥 − 𝑒−𝑖𝑥×

𝑒𝑖𝑥

𝑒𝑖𝑥

𝑧 =𝑖(𝑒𝑖𝑥+𝑖𝑥 − 𝑒−𝑖𝑥+𝑖𝑥)

(𝑒𝑖𝑥+𝑖𝑥 + 𝑒−𝑖𝑥+𝑖𝑥)

𝑧 =𝑖(𝑒2𝑖𝑥 − 𝑒0)

(𝑒2𝑖𝑥 + 𝑒0)

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

8

𝑧 =𝑖(𝑒2𝑖𝑥 − 1)

(𝑒2𝑖𝑥 + 1)

𝑧(𝑒2𝑖𝑥 + 1) = 𝑖(𝑒2𝑖𝑥 − 1)

𝑧𝑒2𝑖𝑥 + 𝑧 = 𝑖𝑒2𝑖𝑥 − 𝑖

𝑧𝑒2𝑖𝑥 − 𝑖𝑒2𝑖𝑥 = −𝑖 − 𝑧

𝑒2𝑖𝑥(𝑧 − 𝑖) = −(𝑧 + 𝑖)

−𝑒2𝑖𝑥(𝑧 − 𝑖) = (𝑖 + 𝑧)

𝑒2𝑖𝑥(𝑖 − 𝑧) = (𝑖 + 𝑧)

𝑒2𝑖𝑥 =(𝑖 + 𝑧)

(𝑖 − 𝑧)

𝑡𝑎𝑘𝑖𝑛𝑔 ln 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒

ln 𝑒2𝑖𝑥 = ln(𝑖 + 𝑧)

(𝑖 − 𝑧)

2𝑖𝑥 ln 𝑒 = ln(𝑖 + 𝑧)

(𝑖 − 𝑧)

2𝑖𝑥 = ln(𝑖 + 𝑧)

(𝑖 − 𝑧)

𝑥 =1

2𝑖ln

(𝑖 + 𝑧)

(𝑖 − 𝑧)

cot−1 𝑧 =1

2𝑖ln

(𝑖 + 𝑧)

(𝑖 − 𝑧)

IV. sec−1 𝑧 =1

𝑖ln

1+√1−𝑧2

𝑧

𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏:

𝑙𝑒𝑡

𝑥 = 𝑠𝑒𝑐−1 𝑧

sec 𝑥 = 𝑧 ⟶ 1

𝑢𝑠𝑖𝑛𝑔

1 + tan2 𝑥 = sec2 𝑥

sec2 𝑥 − 1 = tan2 𝑥

tan 𝑥 = √sec2 𝑥 − 1 ⟶ 2

𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑟𝑜𝑚 1

tan 𝑥 = √𝑧2 − 1

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

9

𝑢𝑠𝑖𝑛𝑔

𝑒𝑖𝑥 = cos 𝑥 + 𝑖 sin 𝑥

𝑒𝑖𝑥 =1

sec 𝑥+ 𝑖

sin 𝑥

cos 𝑥× cos 𝑥

𝑒𝑖𝑥 =1

sec 𝑥+ 𝑖

tan 𝑥

sec 𝑥

𝑒𝑖𝑥 =1 + 𝑖 tan 𝑥

sec 𝑥

𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑟𝑜𝑚 1 𝑎𝑛𝑑 2

𝑒𝑖𝑥 =1 + 𝑖√𝑧2 − 1

𝑧

𝑒𝑖𝑥 =1 + √−1√𝑧2 − 1

𝑧 ∵ 𝑖 = √−1

𝑒𝑖𝑥 =1 + √1 − 𝑧2

𝑧

𝑡𝑎𝑘𝑖𝑛𝑔 ln 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒

ln 𝑒𝑖𝑥 = ln (1 + √1 − 𝑧2

𝑧)

𝑖𝑥 ln 𝑒 = ln (1 + √1 − 𝑧2

𝑧)

𝑥 =1

𝑖ln (

1 + √1 − 𝑧2

𝑧) ∵ ln 𝑒 = 1

sec−1 𝑧 = 1

𝑖ln (

1 + √1 − 𝑧2

𝑧) ∎ ∵ 𝑥 = sec−1 𝑧

V. csc−1 𝑧 =1

𝑖ln (

1+√𝑧2−1

𝑧)

𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏:

𝑙𝑒𝑡

𝑥 = csc−1 𝑧

csc 𝑥 = 𝑧 ⟶ 1

𝑢𝑠𝑖𝑛𝑔

1 + cot2𝑥 = csc2 𝑥

csc2 𝑥 − 1 = cot2𝑥

𝑧2 − 1 = cot2𝑥

cot 𝑥 = √𝑧2 − 1 ⟶ 2

𝑢𝑠𝑖𝑛𝑔

𝑐𝑎𝑙𝑐𝑢𝑙𝑢𝑠

10

𝑒𝑖𝑥 = cos 𝑥 + 𝑖 sin 𝑥

𝑒𝑖𝑥 = sin 𝑥 ×cos 𝑥

sin 𝑥+

𝑖

csc 𝑥

𝑒𝑖𝑥 =cot 𝑥

csc 𝑥+

𝑖

csc 𝑥

𝑒𝑖𝑥 =𝑖 + cot 𝑥

csc 𝑥

𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑟𝑜𝑚 1 𝑎𝑛𝑑 2

𝑒𝑖𝑥 =𝑖 + √𝑧2 − 1

𝑧

𝑡𝑎𝑘𝑖𝑛𝑔 ln 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒

ln 𝑒𝑖𝑥 = ln𝑖 + √𝑧2 − 1

𝑧

𝑖𝑥 𝑙𝑛 𝑒 = ln𝑖 + √𝑧2 − 1

𝑧

𝑥 =1

𝑖ln

𝑖 + √𝑧2 − 1

𝑧 ∵ ln 𝑒 = 1

csc−1 𝑧 =1

𝑖ln

𝑖 + √𝑧2 − 1

𝑧∎ ∵ 𝑥 = csc−1 𝑧